Proving uniform continuity of function:$$f(x,y)=\begin{cases} \frac{x^3-xy}{x^2+y^2}, & (x,y)\neq (0,0) \\ 0, & (x,y)=(0,0) \end{cases}$$
This is supposedly solve, but I don't understand the solution. It is proved that it is continuous and differentiable (asked in sub questions, but might be important for overall idea.) After this, it is proved that $\frac{\partial f}{\partial x}(x,y)\leq 3$ and also $\frac{\partial f}{\partial y}(x,y)\leq 2$. Then it goes onto say: $$|f(x)-f(y)|\leq k \|x-y\|\\ |f(x)-f(y)|\leq |f(x)-f(z)|+|f(z)-f(y)| \leq k \|x-z\|+ k \|z-y\|\\ \leq k(\|x-z\|+ \|z-y\|)...$$
Does this make any sense? Im interested in this variation of the answer...
Forget about uniform continuity. That function is not even continuous at $(0,0)$: It tends to $0$ along the axes and $\to -1/2$ along the line $y=x.$ You can directly check that the partial derivatives are not bounded; for example $\partial f /\partial x (0,y) = -1/y,$ which blows up as $y\to 0.$