Theorem: Let $A ⊂ R^n$ be open and $f: A ⊂ R^n → R^m$. Suppose $f = (f_1,...,f_m)$. If each of the partials $\frac{∂f_j}{∂x_i}$ exists and is continuous on $A$, then $f$ is differentiable on A.
Use the Theorem to show that $f(x,y)$ defined by $f(x,y) = \frac{xy}{\sqrt{x^2+y^2}}$, $(x,y)≠(0,0)$ and $f(x,y)=0, (x,y)=(0,0)$ is differentiable at $(0,0)$
So I first found $\frac{∂f}{∂x}$ = $\frac{xy(2x^2+2y^2-x^2y)}{(x^2+y^2)\sqrt{x^2+y^2}}$ but that is not continuous at $(0,0)$. So I'm not sure how I can prove $f$ is differentiable with this theorem.
The function $$f(x,y) = \frac{xy}{\sqrt{x^2+y^2}}$$ is not differentiable at the origin. If it was, its derivative would be equal to zero as $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$. But then, we would have $$\lim\limits_{(x,y) \to (0,0)} \frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=\lim\limits_{(x,y) \to (0,0)} \frac{x y}{x^2+y^2}=0$$ which is not the case.
However, you can have functions having a derivative at a point while the partial derivatives are not continuous.