Another solution needed

First of all, put $y=0$ , to get $f(0)\cdot f(x)=f(x) \implies f(0)=1$

Now put $x=0$ , to get $f(y)f(-y)=1$ .

Use partial derivatives! Taking the partial derivative with respect to $x$ ,

$f(y)\cdot f'(x-y)=f'(x)$

Putting $y=x \implies f(x)\cdot f'(0)=f'(x) \implies f'(x)=p\cdot f(x)$

Thus $f'(5)=p\cdot f(5) \implies f(5)=\dfrac{q}{p} \implies f(-5)=\dfrac{p}{q}$

Now partial derivative with respect to $y$ ,

$f'(y)f(x-y)-f(y)f'(x-y)=0 \\ \therefore f'(y)f(-y)=f(y)f'(-y)$

$\therefore f'(-y)=\dfrac{f'(y)f(-y)}{f(y)}$

$\therefore f'(-5)=\dfrac{f'(5)f(-5)}{f(5)} = \dfrac{q\times \frac{p}{q}}{\frac{q}{p}}$

$\therefore \boxed{f'(-5)=\dfrac{p^2}{q}}$

differenciate the following equation with respect to x.
f(y).f'(x-y)=f'(x).
Put x=y.
f(x).f'(0)=f'(x).
pf(x)=f'(x).
Take f(x) to the other side and integrate to get.
f(x)=e^(px) +c.
f'(x)=p.e^(px).
So q=p.e^(5p).
e^-(5p)=p/q.
Hence p.e^-(5p)=p^2/q.
f'(-5)=p^2/q