Is $f(x,y)=\frac{x|y|}{\sqrt{x^2+y^2}}$ continuous and differentiable? Let $$f(x,y)=\begin{cases}\frac{x|y|}{\sqrt{x^2+y^2}} &amp;\textrm{ if } (x,y)\neq(0,0), \\ 0 &amp;\textrm{ if } (x,y)= (0,0).\end{cases} $$
I want to know: is $f$ is continuous? Are the partial derivatives exists at the origin? Is $f$ differentiable?
First, I tried some different paths to prove that $f$ is discontinuous, but none path lead me to it.
I found the partial derivatives:
$$\frac{\partial f}{\partial x}(x,y)=\frac{(y^2-x^2)|y|}{(x^2+y^2)^{3/2}},\quad \frac{\partial f}{\partial y}(x,y)=\frac{x|y|(x^2-y^2)}{y(x^2+y^2)^{3/2}}. $$
By doing $(x,y)=(0,t)$, the limit of the first partial derivative does not exist:
$$\lim_{t\rightarrow 0}\frac{t^2|t|}{t^3}=\lim_{t\rightarrow 0}\frac{|t|}{t} $$
Is this sufficient to  conclude that $f$ is not differentiable? What can I do?
Continuity: $(x-|y|)^2=x^2+y^2-2x|y|\geq0 \iff \frac{x^2+y^2}{2}\geq x|y|$

Partial Derivatives: $\frac{\partial f}{\partial x}(0,0)=\lim_{h\to0} \frac{f((0,0)+h(1,0))-f(0,0)}{h}$ and it is analogous for $\frac{\partial f}{\partial y}(0,0)$.

Derivability: Consider the linear application, $L$ , such that for any $(x,y)\in\mathbb{R^2}:L(x,y)=0$ then
$$\lim_{(x,y)\to(0,0)} \frac{f(x,y)-f(0,0)-L((x,y)-(0,0))}{||(x,y)-(0,0)||}=\lim_{(x,y)\to(0,0)} \frac{x|y|}{x^2+y^2}$$


Can you take it from here?
Firstly $f$ is continuous, indeed by polar coordinates we have
$$\frac{x|y|}{\sqrt{x^2+y^2}}=r\cos \theta |\sin \theta| \to 0$$
For differentiability by the definition we have
$$\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$$
and
$$\lim_{(h,k)\to(0,0)} \frac{\frac{h|k|}{\sqrt{h^2+k^2}}}{\sqrt{h^2+k^2}}=\lim_{(h,k)\to(0,0)} \frac{h|k|}{h^2+k^2} \neq 0$$
and therefore $f$ is not differentiable.
As noticed, we cannot conclude for differentiability from the discontinuity of partial derivatives.