$f(x,y) = \frac{x^3y}{x^4 + y^2}$ is not differentiable at $(0,0)$.

Elaborating from the comments to give a complete answer.

The function $f$ would be differentiable at $(0, 0)$ if there existed $(\alpha, \beta) \in \mathbb{R}^2$ such that $$\lim_{(h, k) \to (0, 0)}\dfrac{\vert f(0+h, 0+k) - f(0, 0) - \alpha h - \beta k\vert}{\sqrt{h^2 + k^2}} = 0.$$

From this definition, it also follows that if such a pair $(\alpha, \beta)$ does exist, then $\alpha = f_x(0, 0)$ and $\beta = f_y(0, 0)$. Computing these derivatives (using the limit-definition) is straightforward and gives us $(\alpha, \beta) = (0, 0)$.

Thus, the function is differentiable at $(0, 0)$ if and only if

$$\lim_{(h, k) \to (0, 0)}\dfrac{\vert f(0+h, 0+k) - f(0, 0)\vert}{\sqrt{h^2 + k^2}} = 0$$ $$ \iff \lim_{(h, k) \to (0, 0)}\dfrac{\vert f(h, k)\vert}{\sqrt{h^2 + k^2}} = 0$$

The above is what you had arrived at.
Now, I show that the above limit can not equal $0$ (in fact, it does not exist but we don't need that).

Consider the sequence $(h_n, k_n) = \left(\frac{1}{n}, \frac{1}{n^2}\right)$. It is clear that $(h_n, k_n) \to (0, 0)$ and $(h_n, k_n) \neq (0, 0)$ for any $n \in \mathbb{N}$.

However, note that $$\dfrac{\vert f(h_n, k_n)\vert}{\sqrt{h_n^2 + k_n^2}} = \dfrac{\frac{1/n^5}{2/n^4}}{\sqrt{1/n^2 + 1/n^4}} = \dfrac{1}{2}\dfrac{1}{\sqrt{1 + 1/n^2}} \to \dfrac{1}{2} \neq 0.$$

Thus, by sequential criterion, we get that limit is not $0$. (The limit isn't $\frac{1}{2}$ though.)

Thus, we have shown that the function is not differentiable.

Also, even if you could show that the partial(s) are discontinuous, you wouldn't have proven the non-differentiability. For example, consider the following function $g:\mathbb{R}^2 \to \mathbb{R}.$

$$g(x, y) = \begin{cases} x^2\sin\left(\dfrac{1}{x}\right),& x \neq 0\\ 0,& x = 0 \end{cases}.$$

It is easy to show that the function is indeed differentiable at $(0, 0)$ using the above definition but the partial derivative with respect to $x$ is not continuous.