Show that a complex function $f(x,y)=u(x,y)+iv(x,y)$ with components $u(x,y)=\left\{\begin{matrix} \frac{x^{3}-y^{3}}{x^{2}+y^{2}}& &(x,y)\neq (0,0) \\ 0& &(x,y)=(0,0) \end{matrix}\right.$ and $v(x,y)=\left\{\begin{matrix} \frac{x^{3}+y^{3}}{x^{2}+y^{2}}& &(x,y)\neq (0,0) \\ 0& &(x,y)=(0,0) \end{matrix}\right.$ is not complex differentiable.
At first I tried to see whether the Cauchy-Riemann equations were satisfied. But then I saw my notes and it says in this case, these components satisfy the Riemann-Cauchy equations but yet still $f(x,y)$ is not complex differentiable. (Honestly, I still don't understand this because I calculated $D_{1}u(x,y)=\left\{\begin{matrix} \frac{-(x^{3}-y^{3})(2x)+(x^{2}+y^{2})(3x^{2})}{(x^{2}+y^{2})^{2}}& &(x,y)\neq (0,0) \\ 0& &(x,y)=(0,0) \end{matrix}\right.$ and $D_{2}v(x,y)=\left\{\begin{matrix} \frac{-(x^{3}-y^{3})(2y)+(x^{2}+y^{2})(3y^{2})}{(x^{2}+y^{2})^{2}}& &(x,y)\neq (0,0) \\ 0& &(x,y)=(0,0) \end{matrix}\right.$ and I don't know how they are equal? But I didn't go into the algebraic details and left it)
Do I have to calculate the partial derivatives and make each variable approach $0$ or something? Any tips are appeciated!
You need $u,v$ to be differentiable (as maps $\mathbb{R}^2 \to \mathbb{R}$) in the ordinary sense in an open set, not just that the partials exist.
Take $u$ for example, let $(x(t),y(t)) = t(1,\alpha)$, then ${u(x(t),y(t)) - u(x(0),y(0)) \over t} = {1 - \alpha^3 \over 1+\alpha^2}$, so it is clear that the limit is different depending on $\alpha$. Hence $u$ is not $\mathbb{R}^2 \to \mathbb{R}$ differentiable at $(0,0)$.
(Remember just because ${\partial u(x,y) \over \partial x}$ and ${\partial u(x,y) \over \partial y}$ exist does not mean that $u$ is differentiable. An an extreme example, suppose $u$ is zero everywhere except on the $x,y$ axes where it has value one. Then the partials exist but $u$ is not even continuous.)