Prove the discontinuity of $f $ at origin $ f(x,y)=\begin{cases}\frac{x^2-xy}{x+y},&\text{if $(x,y)\neq(0,0)$} \\ 0,&(x,y) \>\ = (0,0)\end{cases}$ .
Also show that $f $ has partial derivatives at every point including the origin.
I am not able to show discontinuity itself. Because when I substitute $y=mx$ the function is going to zero. Even with the substitution $x = r\cos(\theta),y=r\sin(\theta)$ the function is going to zero . How to prove discontinuity?
I can prove partial derivative existence myself.
As Olof pointed out, this is not well-defined when $y=-x$. Let's ignore this problem and introduce new variables $u=x+y$ and $v=x-y$. Then $$\frac{x^2-xy}{x+y}=\frac{(u+v)v}{2u}=\frac v2+\frac{v^2}{2u}.$$ Approaching the origin on a curve $(u,v)=(t,t)$ gives a limit zero, while on the curve $(u,v)=(t^3,t)$ gives limit infinity. You can translate these paths back into $(x,y)$-coordinates if you so desire.