Hi I have to find the stationary points for $$f(x)= x^4+y^4-(x-y)^2.$$ So far i founded the partial derivatives for $x$ and $y$.
Next step is to solve this system to get my critical points: $$ \left\{ \begin{array}{c} 2x^3-x+y=0 \\ 2y^3-y+x=0 \end{array} \right. $$ But I don't really know how can I solve this system without having to substitute $y=x-2x^3$ and then raise at the power of $3$.
Adding the two equations gives $$2x^3+2y^3=0\\ \implies x^3+y^3=0$$ See here:
http://www4b.wolframalpha.com/Calculate/MSP/MSP3171205d4eh2905ci3e900002931gcaia784ic6i?MSPStoreType=image/gif&s=36&w=300.&h=237.&cdf=Resizeable&cdf=Rotation