auto-correlation of differentiator

How to do this entirely in the time-domain, without using frequency domain?

Let's say I have a continuous-time system that is a differentiator, and X(t) is WSS:

$$x(t) \rightarrow \boxed{\frac{d}{dt}} \rightarrow y(t)$$

I can change the differentiator operator into convolution as follows:

$$X(t) \rightarrow \boxed{\frac{d}{dt} \delta(t)} \rightarrow Y(t)$$

$$h(t) = \frac{d}{dt}\delta(t)$$

$$Y(t) = h(t) * X(t)=\frac{d}{dt}x(t)$$

$$\begin{matrix}R_{XX}(\tau)\end{matrix}\longrightarrow \boxed{\begin{matrix}h(\tau)\end{matrix}}\longrightarrow \begin{matrix}R_{XY}(\tau)\end{matrix}\longrightarrow \boxed{\begin{matrix}h(-\tau)\end{matrix}}\longrightarrow \begin{matrix}R_{YY}(\tau)\end{matrix}$$

cross-correlation of X(t) and Y(t):

$$R_{XY}(\tau) = h(\tau)*R_{XX}(\tau)$$

$$R_{XY}(\tau) = \Big(\frac{d}{d\tau}\delta(\tau)\Big) * R_{XX}(\tau)$$

$$R_{XY}(\tau) = \frac{d}{d\tau} R_{XX}(\tau)$$

autocorrelation of Y(t):

$$R_{YY}(\tau) = h(-\tau)*R_{XY}(\tau)$$

$$R_{YY}(\tau) = h(-\tau)* \frac{d}{d\tau} R_{XX}(\tau)$$

$$h(\tau) = \frac{d}{dt}\delta(\tau)$$

$$h(-\tau) = \frac{d}{dt}\delta(-\tau)$$

$$\delta(\tau) = \delta(-\tau)$$

$$h(-\tau) = \frac{d}{dt}\delta(\tau)$$

$$R_{YY}(\tau) = \Big(\frac{d}{d\tau}\delta(\tau)\Big) * \Big(\frac{d}{d\tau} R_{XX}(\tau)\Big)$$

$$R_{YY}(\tau) = \frac{d^2}{d\tau^2} R_{XX}(\tau)$$

However, the book says the answer should be:

$$R_{YY}(\tau) = -\frac{d^2}{d\tau^2} R_{XX}(\tau)$$

Somewhere I lost a sign...

and the reason why this sign is important because if I take the Fourier of $$R_{YY}(\tau)$$, it needs this sign to make $$S_{YY}(\omega)$$ positive as is required for PSD.

The derivative of the Dirac delta impulse is written as $$\delta'(\tau)$$. This helps with notation because the mistake you made is to write $$h(-\tau)=\frac{d}{d\tau}\delta(-\tau)$$, which is not the case because $$\delta(\tau)$$ is an even (generalized) function, whereas the derivative operator $$\delta'(\tau)$$ is an odd (generalized) function:
$$\delta'(\tau)=-\delta'(-\tau)\tag{1}$$
\begin{align}R_{YY}(\tau)&=R_{XX}(\tau)\star\delta'(\tau)\star\delta'(-\tau)\\&=-R_{XX}(\tau)\star\delta'(\tau)\star\delta'(\tau)\\&=-\frac{d^2}{d\tau^2}R_{XX}(\tau)\tag{2}\end{align}