Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ by $$f(x,y):= \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} &\text{when} \, (x,y) \neq (0,0) \\ 0 &\text{when} \, (x,y)=(0,0)\end{cases}$$ Prove $f$ is $C^1$ on all of $\mathbb{R}^2$ and each of $f_{xx}, f_{xy}, f_{yx}, f_{yy}$ exists at each point of $\mathbb{R}^2$.
I think I'm confused about what it means to be differentiable. Is the derivative going to be a real number or a vector? Also how do I prove $f$ is differentiable without knowing the derivative? I tried to calculate it and it was a huge mess.
For $(x,y)\neq(0,0)$, it is clear that $f$ is differentiable.
For $(x,y) = (0,0)$, recall that $f$ is differentiable if its partial derivatives $f_x, f_y$ are continuous at $(0,0)$.
Using polar coordinates we see that \begin{align*} \lim_{(x,y)\to (0,0)} f_x(x,y) =& \lim_{(x,y)\to(0,0)} \frac{y(x^4-y^4+4x^2y^2)}{(x^2+y^2)^2} \\ =& \lim_{r \downarrow 0} \frac{r^5(\cos^4 \theta - \sin^4 \theta + 4\cos^2 \theta \sin^2 \theta)}{r^4} \\ =& \lim_{r \downarrow 0} r(\cos^4 \theta - \sin^4 \theta + 4\cos^2 \theta \sin^2 \theta) \\ =& 0. \end{align*} So $f_x$ is continuous at $(0,0)$. Similarly for $f_y$. We now conclude that $f$ is differentiable on $\mathbb{R}^2$.
To show that $f_{xx},f_{xy},f_{yx},f_{yy}$ exist at each point on $\mathbb{R}^2$, consider continuity at $(x,y) = (0,0)$, for the case of $(x,y) \neq (0,0)$ is clear by argument.