$$f(x,y)=\begin{cases}\frac{x^2y}{x^4+y^2}&(x,y)\ne(0,0)\\0&(x,y)=(0,0)\end{cases}$$ How can I show, that this function is partially derivable at $(0,0)$, but not continuous there? Regarding partial derivation I used Symbolab, but the result doesn't look promising at all
Regarding continuity, can I substitute $a = x^2$ and $b = y^2$ and then
$f(x,y) = \frac{a\sqrt{b}}{a^2 + b} = \frac{a \cdot (1+\frac{\sqrt{b}}{a})}{a \cdot (a + \frac{b}{a})} = \frac {1+\frac{\sqrt{b}}{a}}{a+\frac{b}{a}} = 0 \text { for lim a } \to \infty $
Hint: Finding if a function is partially derivable at $(0,0)$ means taking the limit of the function like so (the definition of the partial derivative): $$\frac{\partial f}{\partial x}\bigg|_{(x,y)=(0,0)}=\lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}$$ and similarly, $$\frac{\partial f}{\partial y}\bigg|_{(x,y)=(0,0)}=\lim_{h\to0}\frac{f(0,0+h)-f(0,0)}{h}$$
To show it is continuous or not at $(0,0)$, we need $$\lim_{(x,y)\to(0,0)}f(x,y) = f(0,0)$$
If the partial derivatives exist at the origin, but using the formula it isn't continuous there, you've proven what you need to prove
Addendum
You can find the function isn't continuous at the origin by putting $y=kx^2, k\in \Bbb R$ so $$\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^4+y^2} \\ \text{becomes}$$ $$\lim_{x\to0}\frac{x^2\cdot kx^2}{x^4+(kx^2)^2}$$ $$=\lim_{x\to0}\frac{kx^4}{x^4+k^2x^4} \ \text{with} \ x\ne 0$$ $$=\lim_{x\to0}\frac{k}{1+k^2}$$ $$=\frac{k}{1+k^2}$$ which changes for all $k\in \Bbb R$, therefore the limit doesn't exist and the function is not continuous at $(0,0)$ as the limit doesn't equal $f(0,0)=0$