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.
 A: 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 $$t^2-at+b=0$$
A: In the initial equations, make the substitution $x/y=z$. From here $y=x/z$. Then the second equation is $$z+\frac1z=4$$ and the first equation is $$xz+x\frac1{z^2}=6$$
Calculate first $z$ as $$z_{1,2}=2\pm\sqrt3$$Now calculate $x$ then $y$.
A: We are given that $$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$
Consider the expression $(x+y)\left(\frac{x}{y}+\frac{y}{x}\right)$:
$$(x+y)\left(\frac{x}{y}+\frac{y}{x}\right)=x+y+\frac{x^2}{y}+\frac{y^2}{x}$$
so
$$(x+y)\left(\frac{x}{y}+\frac{y}{x}-1\right)=\frac{x^2}{y}+\frac{y^2}{x}$$
But using the initial conditions we see that
$$3(x+y)=6\implies x+y=2\implies x=y-2$$
which can be subsituted into your equations to provide solutions for $x$ and $y$.
This is a bit more of an experimental solution than the other answers; the expression considered intially is not necessarily obvious. Often though some fiddling around with expressions like that is useful.
I hope that helps :)
