Solving the given system of equations 
$$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$

I got $x^3 + y^3 = 6xy = (x+y)^2$ which I simplified to $x^2 - xy + y^2 = x + y$ and I'm not sure what to do now. Please help.

Hint:
Let $x+y=a, xy=b$
$$\implies 6b=a^3-3ab\  \ \ \  (1)\text{ and }4b=a^2-2b\implies a^2=6b\  \ \ \  (2)$$
Clearly, $ab\ne0$
Replacing the value of $b$ in $(1)$,
$$a^2=a^3-3\cdot a\cdot\dfrac{a^2}6\iff2a^2=a^3\implies a=?\text{ as }a\ne0$$
So, we can find $b=\dfrac{a^2}6$
So, $x,y$ are the roots of $$t^2-at+b=0$$
