Question on solving partial derivative in probability theory 
When the diffusion process $X_{t}$ is stationary, $F_{s, x}(s+t, y)$ does not depend on $s$. Hence we have a well-defined function
$$
F_{t}(x, y)=F_{s, x}(s+t, y), \quad t>0 .
$$
Since $F_{s, x}(s+t, y)$ does not depend on $s, \frac{d}{d s} F_{s, x}(s+t, y)=0$. This implies the second equality below:
$$
\frac{\partial}{\partial t} F_{t}\left(x, y_{0}\right)=\frac{\partial}{\partial(s+t)} F_{s, x}\left(s+t, y_{0}\right)=\left.\left(-\frac{\partial}{\partial s} F_{s, x}\left(u, y_{0}\right)\right)\right|_{u=s+t} .
$$
Then use Equation (10.9.8) to get
$$
\frac{\partial}{\partial t} F_{t}\left(x, y_{0}\right)=\rho(x) \frac{\partial}{\partial x} F_{t}\left(x, y_{0}\right)+\frac{1}{2} Q(x) \frac{\partial^{2}}{\partial x^{2}} F_{t}\left(x, y_{0}\right)
$$

I am reading a textbook and see this part. I wonder how do I get the second equality as mentioned here.

$$\frac{d}{ds}F_{s,x}(s+t,y)=0$$
$$\Rightarrow \frac{\partial F_{s,x}(u,y)}{\partial s}\lvert_{u=s+t}+\frac{\partial F_{s,x}(u,y)}{\partial u}\frac{\partial u}{\partial s}=0$$
\begin{align}
\frac{\partial F_{s,x}(u,y)}{\partial s}\lvert_{u=s+t}
&= -\frac{\partial F_{s,x}(u,y)}{\partial u}\frac{\partial u}{\partial s} \\
&= -\frac{\partial F_{s,x}(u,y)}{\partial s}*1 \\
&= -\frac{\partial F_{s,x}(u,y)}{\partial s}
\end{align}
