A function $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable at a point $p$ if there is a linear map $L:\mathbb{R}^2 \to \mathbb{R}$ such that $$\lim_{h \to 0}\frac{\left|f(p+h)-f(p)-L(h)\right|}{|h|} = 0$$ If the function is differentiable, then $L$ is the function $L(x,y) = \dfrac{\partial f }{\partial x}\big|_p x +\dfrac{\partial f}{\partial y}\big|_p y$.
If all you know is that the partial derivatives exist, you do not even know that the function is continuous, let alone is well approximated by a tangent plane. For instance $f(x,y) = 0$ if $x=0$ or $y=0$ and $f(x,y) = 1$ otherwise. The partial derivatives of $f$ at $(0,0)$ are all $0$, but the tangent plane is a really crappy approximation to $f$ off of the coordinate axes.
The example you were looking at is a little harder to visualize, but think about what happens to $f$ on lines other than horizontal and vertical ones.
In other words, morally, for a function to be differentiable at $(a,b)$ we need that $f(a+\Delta x,b+\Delta y) \approx f(a,b) + \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y$. This talks about approximating $f$ in all directions around $(a,b)$, whereas existence of the partial derivatives only means you have good approximations in two directions (the coordinate directions).
You are on the right track but the question is, how to compute the following in general :
$$\lim_{(x,y)\to(a,b)}h(x,y).$$
We may not just do $x \to a$ then $y \to b$ (or the converse) for example. We need to consider all the manners for $(x,y)$ to tend to $(a,b)$.
Example of why it does not work :
Consider $f(x,y)= \chi_{\{x=0\} \cup \{y=0\}}$. With $\chi_A$ the function whose value is $1$ on $A$ and $0$ otherwise. Then $f$ is not continuous at $0$ but :
$$\lim_{x \to 0} f(x,0)=1=\lim_{y \to 0} f(0,y).$$
But obviously the limit for both variables does not exists, since :
$$\lim_{t \to 0} f(t,t)=0.$$
Which brings us to one of the method to show a function is not continuous at a point. Just find two different manners to tend to the point of interest such that the limit is different.
In your case the answer is clear since the partial derivative you found are polynomials.
But one way (the most usual one I think) is to majorate the quantity $|h(a+u,b+v)-h(a,b)|$ by something who obviously tends to $0$ when $(u,v) \to 0$ or which had been proved to tend to $0$.
Finally, here is you want to prove that the function are continuous, i.e. if you consider not known that polynomials are continuous, you have to get back to definition of continuity with $\varepsilon, \delta$.
Let $\varepsilon >0$, suppose $\|(u,v)-(a,b)\|_1 \le \delta$ with $\delta = \varepsilon / 2$ then :
$$|f_x(u,v)-f_x(a,b)|=|2u+v-(2a+b)|\le 2|u-a| + |v-b| \le 2 \|(u,v)-(a,b)\|_1 \le \varepsilon$$
Then $f_x$ is continuous at $(a,b)$. I let you try for $f_y$.
Best Answer
The differentiability theorem states that continuous partial derivatives are sufficient for a function to be differentiable. It's important to recognize, however, that the differentiability theorem does not allow you to make any conclusions just from the fact that a function has discontinuous partial derivatives. The converse of the differentiability theorem is not true. It is possible for a differentiable function to have discontinuous partial derivatives.
An example of such a strange function is
$$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & \text{ if $(x,y) \ne (0,0)$}\\0 & \text{ if $(x,y) = (0,0)$}.\end{cases}$$
Since the argument of the sinusoid approaches infinity as one approaches the origin, it oscillates wildy near the origin. But, the sinusoid is bounded between $−1$ and $1$, and the oscillations of the sinusoid are tempered by the quadratic term $x^2+y^2$. As both paraboloids have the same horizontal tangent plane at the origin $z=0$, we can infer that $f(x,y)$ has the same tangent plane at the origin. The function $f(x,y)$ is differentiable at the origin.
To verify that the partial derivatives of $f$ are zero at the origin, consistent with the horizontal tangent plane, one must use the limit definition. For example, the derivative with respect to $x$ can be calculated by
\begin{align*} f_x(0,0) &= \lim_{h \to 0} \frac{f(h,0)-f(0,0)}{h}\\ &= \lim_{h \to 0}\frac{h^2 \sin (1/|h|)-0}{h} \\ &= \lim_{h \to 0}h \sin (1/|h|) =0. \end{align*} A similar calculation shows that $f_y(0,0)=0$. Away from the origin, one can use the standard differentiation formulas to calculate that \begin{align*} f_x(x,y) &= 2 x \sin \left(\frac{1}{\sqrt{x^2+y^2}}\right)-\frac{x \cos \left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}}\\ f_y(x,y) &= 2 y \sin \left(\frac{1}{\sqrt{x^2+y^2}}\right)-\frac{y \cos \left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}}. \end{align*} Both of these derivatives oscillate wildly near the origin. For example, the derivative with respect to $x$ along the $x$-axis is
$$f_x(x,0) = 2 x \sin \left(1/|x|\right)-\text{sign}(x) \cos \left(1/|x|\right),$$
for $x≠0$, where $\text{sign}(x)$ is $±1$ depending on the sign of $x$. In this case, the sine term goes to zero near the origin but the cosine term oscillates rapidly between $1$ and $−1$, as it is not multiplied by anything small. As $f_x$ approaches both $1$ and $-1$ within any neighborhood of the origin, it is discontinuous there. In the same way, one can show that $f_y$ has wild oscillations and is discontinuous at the origin. This function should prevent you from misinterpreting the differentiability theorem, as you now know that continuous partial derivatives imply differentiability but not vice-versa.