Short Answer: $16x^2+9y^2=144z^2$
Derivation:
The little trick here is to exploit the fact that $z(u,v)=u$. Our surface is given by
$$(x(u,v),y(u,v),z(u,v)) = (3u\cos v,4u\sin v,u).$$
We know that $\cos^2v+\sin^2v=1$ for all $v$ and we can use this fact. Notice that:
$$\frac{x}{3z} = \frac{3u\cos v}{3u} = \cos v$$
Similarily, we can show that $y/4z = \sin v$ and then use the identity $\cos^2v = \sin^2v\equiv 1$:
$$\left( \frac{x}{3z} \right)^{\! 2} + \left( \frac{y}{4z} \right)^{\! 2} = 1$$
Expanding out all of the powers gives the following chain of events:
$$\left( \frac{x}{3z} \right)^{\! 2} + \left( \frac{y}{4z} \right)^{\! 2} = 1 \implies \frac{x^2}{9z^2}+\frac{y^2}{16z^2}=1 \implies 16x^2+9y^2=144z^2$$
We can check this equation just to make sure it's correct:
$$16x^2+9y^2=16(3u\cos v)^2+9(4u\sin v)^2 = 144u^2\cos^2v + 144u^2\sin^2v=144u^2 = 144z^2$$
GENERAL WARNING
Let $P$ be the set of points given by a parametrisation, and let $E$ be the set of points given by an equation. We have shown that every point in $P$ is a point in $E$, that is to say $P$ is a subset of $E$. It might be the case that there are points in $E$ that the parametrisation does not cover. Think of $P$ as $(x(t),y(t)) = (t,\sqrt{t})$ and $E$ as $x=y^2$. (The points with $x=y^2$ and $y<0$ are not in $P$ although they are in $E$.) To show that an equation and a parametrisation give the same set of points we must show that all the points of the equation's solution are covered by the parametrisation ($E \subseteq P$) and that all of the points of the parametrisation satisfy the equation ($P \subseteq E$). If $P \subseteq E$ and $E \subseteq P$ then $E=P$, just like with ordinary numbers: if $a \le b$ and $b \le a$ then $a=b$.
The usual way to solve this type of problem, once you’ve verified that the vector field is in fact conservative, is to alternate integration and differentiation.
We want to find some scalar function $f$ such that $f_x=4y\sin(xy)$ and $f_y=4x\sin(xy)$, so start by integrating one of the components, say, the first one: $$f(x,y) = \int 4y\sin(xy)\,dx = -4\cos(xy)+g(y).\tag{1}$$ Notice that instead of being a simple constant as it would be for functions of one variable, the “constant” of integration is some unknown function that doesn’t depend on $x$. Now, differentiate the result with respect to $y$ and compare this to the corresponding component of the vector field: $${\partial\over\partial y}f(x,y)=4x\sin(xy)+g'(y)=4x\sin(xy).$$ This implies that $g'(y)=0$, so in this case the “constant” of integration in (1) really is just a constant and $f(x,y)=-4\cos(xy)+C$. Otherwise, we’d continue by integrating $g'(y)$ with respect to $y$ to get $f$. For functions of more than two variables, you would go back and forth like this, alternately integrating and differentiating with respect to each variable in turn.
It’s possible to find a scalar field $f$ all at once, too. Let $F(x,y)$ be the vector field and define $f(x,y)=\int_\Gamma F$, where $\Gamma$ is some smooth curve joining the origin to $(x,y)$. Since $F$ is conservative, the value of this integral depends only on the endpoints of $\Gamma$, so $f$ is well-defined. It’s easy to see that with this definition, $F=\nabla f$. To evaluate this integral, choose any convenient path, such as the line segment from the origin to $(x,y)$ parametrized by $\gamma:t\mapsto(tx,ty)$, $0\le t\le1$, which results in the ordinary integral $$f(x,y)=\int_0^1F(tx,ty)\cdot(x,y)\,dt.$$ For this problem, we have $$\begin{align} f(x,y)&=\int_0^14xyt\sin(xyt^2)+4xyt\sin(xyt^2)\,dt \\
&= 4\int_0^1 2xyt\sin(xyt^2)\,dt \\
&= -4\cos(xy)+4. \end{align}$$ This differs from the above result by a constant, which of course is arbitrary.
Best Answer
To undo, you need to play a guessing game: Note that if $\vec{v} = \vec{\nabla}f$, then for your case, we have $$\dfrac{\partial f}{\partial x} = 6 \cos(x^2+4y^2) - 12x^2\sin(x^2+4y^2) \tag{$\star$}$$ and $$\dfrac{\partial f}{\partial y} = - 48xy\sin(x^2+4y^2)\tag{$\dagger$}$$ It is easier to deal with $\dagger$ than $\star$. Note that $$\dfrac{\partial f}{\partial y} = 6x\dfrac{d(y^2)}{dy}\dfrac{d\left(\cos(x^2+4y^2)\right)}{d(y^2)} = \dfrac{\partial}{\partial y}\left(6x \cos(x^2+4y^2)\right)$$ Hence, $$f(x,y) = 6x \cos(x^2+4y^2) + g(x)$$ Now plug in $f(x,y)$ into $\star$; we then get that $g(x) = \text{constant}$. Hence, $$f(x,y) = 6x \cos(x^2+4y^2) + \text{constant}$$