So I have this function: $f(x,y)=\frac{xy(x^2-y^2)}{(x^2+y^2)}$ for $(x,y)\neq(0,0)$ and $f(x)=0$ for $(x,y)=(0,0)$
I am asked to show:
1) $\frac{\partial{f}}{\partial{x}}(0,0)=\frac{\partial{f}}{\partial{y}}(0,0)=0$
2) $\frac{\partial^2{f}}{\partial{y^2}}(0,0) = 1$ and $\frac{\partial^2{f}}{\partial{y^2}}(0,0) = -1$
3) Explain why the mixed partial derivatives are not equal
I have computed the first partial derivatives for $f(x)$, and I know that I need to use limit to show these partial derivatives' values at $(0,0)$, but I don't know how exactly I should approach them.
For 1) I get $$\frac{\partial{f}}{\partial x}= \frac{3yx^2-y^3}{x^2+y^2}-\frac{2x^4y-2x^2y^3}{\left(x^2+y^2\right)^2}, \\ \frac{\partial f}{\partial y}=\frac{-3xy^2+x^3}{x^2+y^2}-\frac{2x^3y-2xy^3}{\left(x^2+y^2\right)^2},$$ so we will need to evaluate following limits: $$\lim_{\substack{\left(x,y\right)\rightarrow \left(0,0\right) \\ x,y\neq 0}}\left[\frac{3yx^2-y^3}{x^2+y^2}-\frac{2x^4y-2x^2y^3}{\left(x^2+y^2\right)^2}\right], \\ \lim_{\substack{\left(x,y\right)\rightarrow \left(0,0\right) \\ x,y\neq 0}}\left[\frac{-3xy^2+x^3}{x^2+y^2}-\frac{2x^3y-2xy^3}{\left(x^2+y^2\right)^2}\right],$$ and according to the question show that they both $=0$. The first limit simplifies a bit to $$ \lim_{\substack{\left(x,y\right)\rightarrow\left(0,0\right) \\ x,y\neq 0}}\frac{-4y^3}{x^2+y^2}+\lim_{\substack{\left(x,y\right)\rightarrow\left(0,0\right) \\ x,y\neq 0}}\frac{6x^2y^3+2y^5}{y^4+2x^2y+x^4}, $$ and the second goes to $$ \lim_{\substack{\left(x,y\right)\rightarrow\left(0,0\right) \\ x,y\neq 0}}\frac{-4y^2}{y^2+x^2}+\lim_{\substack{\left(x,y\right)\rightarrow \left(0,0\right) \\ x,y\neq 0}}\frac{2xy^3-2x^3y}{\left(x^2+y^2\right)^2}, $$ and now let just $y$ go to $0$ and observe that all of the limits will then be $0$, regardless of what $x$ is.