Solve the equation: $x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$ Here is the equation:

$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$$

The equation looks very simple.  But, it's a little misleading (for me).
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}\Longrightarrow x(x^2-b^2)-a^2x(x^2-b^2)+b^2+a(x^2-b^2)-x^2=0\Longrightarrow (x^2-b^2)(x-a^2x+a-1)=0$$
We have  $x(1-a^2)=1-a$ and $x≠±b$.

1) $a=1$, $x\in (-\infty;b) ∪(-b;b)∪(b;\infty)$
2) $a=-1$ then $x\in\emptyset$
3) $a≠±1$ and $b≠±\frac{1}{a+1}$ ,then $x=\frac{1}{a+1}$

What I want to know, is there a mistake in the solution, or is there any other point I missed?
Thank you.

Your solution is almost right, except that 2) is not a solution because that yields $0=2$ in your equation.
Here’s a simpler method.
Adding $\frac{b^2}{b^2-x^2}$ to both sides, note that the RHS now becomes $1$. So we now need to solve $x(1-a^2)=1-a$, which factorises to $(1-a)(x(a+1)-1)=0$ (where $x \neq \pm b$).
