Partial Differentiation of $\frac 00$

Question

Let: $$f(x,y)=x^2y\sin\left(\frac{y}{x}\right),\ x\neq0$$ $$f(x,y)=0, \ x=0$$ Partial differentiation is obvious for $x\neq0$, however, for $x = 0$ and the derivative over $x$, one gets: $$\lim_{h\to 0}\frac{f(h,y)}{h}$$ which is not constant $= 0$. Do I need to use L'Hospital? If yes differentiating over which variable? I'm new to this topic.

Answer 1

At $x = 0$, the definition of partial derivative gives us $$ \frac{\partial f}{\partial x}(0, y) = \lim_{h\to 0}\frac{f(h, y) - f(0, y)}{h}\\ = \lim_{h\to 0}\frac{h^2y\sin(y/h) - 0}{h}\\ = \lim_{h\to 0}hy\sin(y/h) = 0 $$

Answer 2

$\lim_{h\to0}\frac{f(h,y)}h=\lim_{h\to0}hy\sin\frac yh=0$ since $y$ is (constant) finite and $\sin\frac yh$ is bounded.