Your very close to the correct answer. When you compute the integrals the 'constants' are constant with respect to the variable of integration. For example, your first integral should be:
$$v=\int 3x^2y-y^3 dy = \frac{3}{2}x^2y^2-\frac14y^4 + \phi(x)$$
where $\phi(x)$ is a function of $x$ to be determined.
Now differentiating with respect to $x$, then using the Cauchy-Riemann equations gives:
$$\frac{\partial v}{\partial x} = 3xy^2+\phi'(x) \quad \Rightarrow \quad \phi'(x) = -\frac{\partial u}{\partial y} - 3xy^2 = -x^3$$
So that $\phi(x) = -\frac14 x^4+C$, for constant $C$. Thus we have:
$$v=\frac{3}{2}x^2y^2-\frac14y^4-\frac14x^4 + C$$
Notice that this means that we can only determine $f$ up to a constant.
Here is one way to make the connection you are hinting at more rigorous. Throughout this answer, $\log$ denotes only the real logarithm (defined for positive real numbers only, so there is no ambiguity in how it is interpreted).
Recall that
$$
|e^w| = e^{\operatorname{Re} w}, \qquad w \in \mathbb{C}.
$$
If there were an annulus $A$ centered at $0$ and an function $g$ on $A$ with the property that
$$
\operatorname{Re} g(z) = \log |z|, \qquad z \in A,
$$
then the function $h$ on $A$ given by $h(z) = e^{g(z)}/z$ for all $z \in A$ would be analytic on $A$ and would satisfy
$$
|h(z)| = \frac{|e^{g(z)}|}{|z|} = \frac{e^{\operatorname{Re} g(z)}}{|z|} = \frac{e^{\log |z|}}{|z|} = 1, \qquad z \in A.
$$
It would follow from this that
$$
\overline{h(z)} = \frac{|h(z)|^2}{h(z)} = \frac{1}{h(z)},
$$
being a quotient of analytic functions, is also analytic on $A$.
It easily follows from the Cauchy-Riemann equations that $h$ must be constant on $A$ (generally, whenever both a function and its complex conjugate are analytic on a connected open set, the function must be constant on that set).
So there is $\omega \in \mathbb{C}$ with $|\omega| = 1$ and
$$
e^{g(z)} = \omega z, \qquad z \in A.
$$
Choosing a real number $s$ with $\omega = e^{is}$ it follows that the function analytic $L$ on $A$ given by $L(z) = e^{g(z) - is}$ satisfies
$$
e^{L(z)} = z, \qquad z \in A.
$$
To summarize: beginning with an analytic function having $\log |z|$ as its real part on $A$, we have constructed an analytic branch of the logarithm on $A$. So if we happen to know that no such thing exists, we have our contradiction, and are done.
It is, of course, generally true that if $g$ and $k$ are analytic functions on a connected open set $G$ and $\operatorname{Re} g(z) = \operatorname{Re} k(z)$ holds for all $z \in G$, then there must be a real constant $C$ with $g(z) = k(z) + iC$. (Proof: consider the Cauchy-Riemann equations for the difference $g - k$ to deduce that $g - k$ must be constant, and clearly the constant must have zero real part.) So the intuition that an analytic function having $\log |z|$ as its real part must be "essentially the same thing" as a branch of the logarithm is correct. But in translating this intuition into a nonexistence proof, one must be careful, as you probably sensed when you were asking this question. (Working hastily and without careful thought, one might, for example, try to prove what you want by applying the result just mentioned with $G$ equal to the annulus, and one of the functions $g$ or $k$ equal to a branch of the logarithm--- conveniently forgetting for a moment that the result being appealed to concerns analytic functions on $G$, and there isn't an analytic logarithm on the annulus.)
To get an actual proof along these lines, one must proceed more carefully: let $A$ be the annulus, and let $k(z)$ denote e.g. the principal branch of the logarithm. Then $k$ is analytic on the connected open set $G = A \setminus (-\infty,0]$. If $g$ is an analytic function on $A$ satisfying $\operatorname{Re} g(z) = \log |z|$ for all $z \in A$, the result of the previous paragraph implies that there is a real constant $C$ with the property that $g(z) = k(z) + iC$ holds for all $z \in G$ (not all $z \in A$). Hence $k(z) = g(z) - iC$ holds for all $z \in G$, and as $g$ is continuous on all of $A$, we deduce from this equation that for any $c$ in the nonempty set $A \cap (-\infty,0)$, the limit $\lim_{z \to c, z \in G} k(z)$ exists (and is $g(c) - iC$). This contradicts the fact, obvious from the explicit formulas for $k$, that $k$ has a jump discontinuity at every point on the negative real axis.
Best Answer
Even though in principle the Cauchy-Riemann equations can function as differential equations that will allow you to reconstruct a function in this way, in practice doing this directly can be extraordinarily difficult. Exercises of this kind are almost always meant to be solved using a combination of inspired guesswork and previous knowledge.
If you know of the complex exponential and logarithm (and if you don't you certainly have some work cut out for you here), you should immediately notice that the imaginary part you're expected to find here is exactly the real part of the complex logarithm. So we can very quickly manufacture a function that satisfies it, simply by multiplying by $i$: $$ f(z) = i\operatorname{Log} z $$
The only problem with this is that it is not continuous (and thus not analytic) along the cut, wherever your definition of Log places the principal cut. So the next step would be to figure out whether this is unavoidable.
It turns out that this problem cannot be avoided. Clearly if we're given the imaginary part of the function, we can add any real number we want to the entire function without losing analyticity -- so we can arbitrarily decide that the real part of $f(1)$ is going to be $0$.
But then the Cauchy-Riemann equations immediately tell that the real part of $f(x)$ for any positive real $x$ must be $0$ to, simply because $\frac{du}{dx}=\pm \frac{dv}{dy}$ (I don't even bother to look up the sign), and $\frac{dv}{dy}$ is clearly $0$ everywhere on the real axis, given the specified for $v$.
Thus, we now know the entire complex value of $f$ everywhere on the positive real axis. And we can then use a theorem that says that knowing the value of an analytic function on just a small (nontrivial) piece of curve going through its domain determines its values everywhere. So since $i \operatorname{Log} z$ matches the specified values on the positive real axis, it has to match everywhere, at least if the domain of $f$ includes the domain of $\operatorname{Log}$.
Otherwise the domain of $f$ can be stranger, and we then know only that $f(z)$ must be $i$ times some logarithm of $z$ at each point -- but we still cannot make an $f$ that's continuous all the way around the origin, because the argument (which must become minus the real part of $f$) cannot match up.