In BS world, we have the stock process in log space $dS_t=(r-\frac{1}{2}\sigma^2)dt+\sigma dW$. Let's say we want to price $f(t,x)=\mathbb{E}_{t,x}[h(S(T)]$. Using Feynman-Kac, we get \begin{equation} \frac{\partial f}{\partial t} + (r-\frac{1}{2}\sigma^2)\frac{\partial f}{\partial x}+\frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial x^2}-rV=0 \end{equation}
On the other hand, if we consider the forward process (again in log space) $F_t=S_t+r(T-t)$, we have the forward process $dF_t=-\frac{1}{2}\sigma^2 dt+\sigma dW$ and the price becomes $f(t,y)=\mathbb{E}_{t,y}[h(F(T)]$. Using F-K again, we get \begin{equation} \frac{\partial f}{\partial t} - \frac{1}{2}\sigma^2\frac{\partial f}{\partial y}+\frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial y^2}-rV=0 \end{equation}
Somehow I fail to transform the first PDE to the second by change of variable directly from $S_t$ to $F_t$. Since $y=x+r(T-t)$, by chain rule, $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial y}{\partial x}=\frac{\partial f}{\partial y}$, i.e., the first order is the same and so as the second order. So I end up with \begin{equation} \frac{\partial f}{\partial t} +(r-\frac{1}{2}\sigma^2)\frac{\partial f}{\partial y}+\frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial y^2}-rV=0 \end{equation} which is obviously wrong and I couldn't figure out why.
If $$ f_{X}(t,x)=f_{Y}(t,y(t,x)) $$ where $$ y(t,x)=x+r(T-t) $$ then $$ \frac{\partial f_{X}}{\partial t}= \frac{\partial f_{Y}}{\partial t} + \frac{\partial f_{Y}}{\partial y}\frac{\partial y}{\partial t}= \frac{\partial f_{Y}}{\partial t} - r\frac{\partial f_{Y}}{\partial y}. $$