$\frac{x^{2}}{y}+\frac{y^{2}}{x}=a$ and $\frac{x}{y}+\frac{y}{x}=b ; x$ and $y$ which values satisfy the two equations?
The given equations can also be written as:

$$
x^{3}+y^{3}=a x y
$$

$$
x^{2}+y^{2}=b x y
$$

(1) can also be written as:

$$
(x+y)\left(x^{2}+y^{2}-x y\right)=a x y
$$

Substituting (2) into this:

$$
(x+y) x y(b-1)=a x y
$$

This equation is satisfied by the values $x=y=0$. Dividing by $x y$:

$$
x+y=\frac{a}{b-1}
$$

Subtracting (2) from the square of (4):

$$
2 x y=\frac{a^{2}}{(b-1)^{2}}-b x y
$$

From which we get:

$$
x y=\frac{a^{2}}{(b-1)^{2}(b+2)}
$$

Thus, $x$ and $y$ are, based on (4) and (5), the roots of the following equation:

$$
u^{2}-\frac{a}{(b-1)} u+\frac{a^{2}}{(b-1)^{2}(b+2)}=0
$$

From which we get:

$$
u_{1}=x_{1}=y_{2}=\frac{a\left(b+2+\sqrt{b^{2}-4}\right)}{2(b-1)(b+2)}
$$

and

$$
u_{2}=x_{2}=y_{1}=\frac{a\left(b+2-\sqrt{b^{2}-4}\right)}{2(b-1)(b+2)}
$$

Number of solutions: 27.