Power spectral density of $\left(x(t)\right)^2$?
Since the question has been raised as to whether the hint that I had given to the OP in a comment on the original question was appropriate for a newcomer to signal processing, here goes.

Stripped of extraneous baggage and notation, the question is whether it is possible to determine the value of $E[X^2Y^2]$ straightforwardly where   $X$ and $Y$ are zero-mean unit variance jointly Gaussian random variables with correlation coefficient $\rho$, or is it necessary to resort
to advanced methods or results such as the one cited in the answer by @MattL or generalizations thereof such as Price's Theorem (cf. the
book cited in MattL's answer).  One straightforward method involves
conditional expectations and the result that is known to be a LIE:

$$E[X^2Y^2] = E\big[E[X^2Y^2\mid Y]\big]$$

Sorry, I couldn't resist: LIE is an acronym for the Law of Iterated Expectation which asserts that the expectation of $A$ is the
same as the expectation of the random variable $E[A\mid B]$. Note that $E[A\mid B]$ is a function of $B$, not $A$ as one might naively believe. We already squeezed out all of the randomness in $A$ when we took its
expectation. The LIE is
that the expected value of $E[A\mid B]$, a function of $B$) works out, by a miracle of
modern mathematics, to be the same as the expected value of $A$.

Now, for any choice of real number $y$, given that $Y=y$,

$$E[X^2Y^2\mid Y = y] = E[X^2y^2\mid Y = y]
= y^2 E[X^2\mid Y = y].$$
But, given that $Y = y$, we know that the conditional distribution
of $X$ given $Y=y$ is a Gaussian distribution with mean $\rho y$
and variance $1-\rho^2$. Consequently,
$$E[X^2\mid Y = y] = \operatorname{var}(X\mid Y=y) 
+ \left(E[X\mid Y = y]\right)^2 = (1-\rho^2) + (\rho y)^2$$
leading to
$$E[X^2Y^2\mid Y = y] = y^2((1-\rho^2) + (\rho y)^2))
= y^2(1-\rho^2) + \rho^2y^4 $$
giving us that the random variable $E[X^2Y^2\mid Y]$ is just
$Y^2(1-\rho^2) + \rho^2Y^4$. Since $E[Y^2] = 1$ and $E[Y^4] =  3$,
we get that 
$$E[X^2Y^2] = E\big[E[X^2Y^2\mid Y]\big] = E[Y^2(1-\rho^2) + \rho^2Y^4]
= (1-\rho^2) + 3\rho^2 = 1 +2\rho^2.$$



Finally, noting that $\rho = e^{-a|t|}$, we can write
$$R_y(t) = 1 + 2e^{-2a|t|}$$
just as has been explained in MattL's answer.