Mean square derivative of square of WSS (wide sense stationary) Gaussian RP

Question

Question

Let $X_t$ be a WSS (wide sense stationary) gaussian random process. If $Y_t=X_t^2$ then what is

1. $Y_t'$
2. Covariance of $Y_t'$ in terms of the variance the auto correlation $R_{XX}$

Note that the derivative is the mean square derivative (i.e in $L^2$)

Attempt

So I think for $1$ its $2X_tX'_t$. I have to show that $$E\left(\frac{Y_{t+h}-Y_t}{h}-2X_tX_t'\right)^2 \rightarrow 0.$$

Thus \begin{align*} E\left(\frac{X^2_{t+h}-X^2_t}{h}-2X_tX_t'\right)^2&=E\left(\frac{1}{h^2}( X^4_{t+h}-2X^2_{t+h}X^2_t+X_t^2)-\frac{4}{h}(X_{t+h}^2-X_t^2)(2X_tX_t')+4X_t^2X_t'^2\right). \end{align*}

Now I attempted to split this up using linearity and then I tried to use the fact that WSS implies strict sense stationary (SSS) for Gaussian RP's. However I think I am not properly controlling the fourth moments so I am unable to get the answer I want.

I suspect I can do $2$ after I properly do $1$.