Find the stationary points: $$f(x,y)=x^2-\frac 13y^3-xy$$

Have so far with respect for $x$: $\frac{\partial f}{\partial x}=2x-y$

With respect for $y$: $\frac{\partial f}{\partial y}=-y^2-x$

But having trouble getting from here?


The stationary points are the solutions of the two equations: $$2x-y=0$$ $$-y^2-x=0$$

that is

$$y=2x$$ and $$-4x^2-x=0$$

i.e., $x=0$ or $x=-1/4$

That is, the stationary points are $(0,0)$ and $(-1/4,-1/2)$


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