Find all nonzero real numbers $x$ such that

$$
x^{2}+\frac{36}{x^{2}}=13 \text {. }
$$
Multiplying through by $x^{2}$ and moving all terms to the left gives

$$
x^{4}-13 x^{2}+36=0 \text {. }
$$

We can factor this as

$$
\left(x^{2}-4\right)\left(x^{2}-9\right)=(x-2)(x+2)(x-3)(x+3)=0 .
$$

Thus, the solutions are $x= \pm 2$ and $x= \pm 3$.