How to solve this difficult system of equations? $$1+4\lambda x^{3}-4\lambda y = 0$$
$$4\lambda y^{3}-4\lambda x = 0$$
$$x^{4}+y^{4}-4xy = 0$$
I can't deal with it. How to solve this?

As noted in several comments, the second equation yields $x=y^{3}$. Hence, by the third equation, $$y^{12}=3y^{4} \implies y^{8}=3$$
Assuming $y$ is real, we get $y=\pm 3^{1/8}$ and $x=\pm3^{3/8}$. By the first equation, then, (taking the positive roots): $$1+4\lambda3^{9/8}-4\lambda3^{1/8}=0 \implies 8\lambda3^{1/8}=-1$$
So $\lambda=\frac{-1}{8}3^{-1/8}$. If we were to take the negative roots, $\lambda$ would be positive.
NB: Notice that none of $x,y,\lambda$ can be zero due to equation $1$.
