Really now. \begin{align*}\frac{1}{f(n)}+\frac{1}{f(m)}=\frac{4}{f(n)+f(m)} &\implies \frac{f(n)+f(m)}{f(n)f(m)}=\frac{4}{f(n)+f(m)} \\ &\implies f(n)^2+f(m)^2+2f(n)f(m)=4f(n)f(m) \\ &\implies f(n)^2+f(m)^2-2f(n)f(m)=\left(f(n)-f(m)\right)^2=0 \\ &\implies f(n)-f(m)=0 \\ &\implies f(n)=f(m)\implies n=m\end{align*}as desired.

We have \[\frac{f(n)+f(m)}{f(m)f(n)}=\frac{4}{f(n)+f(m)}\implies f(n)^2+f(m)^2+2f(m)f(n)=4f(m)f(n)\implies f(n)^2+f(m)^2-2f(m)f(n)=(f(n)-f(m))^2=0\implies f(m)=f(n)\implies m=n\]

If $m = n$, then the condition clearly holds for all $f(m) = f(n) \ne 0$. Now, we will show this is the only possibility. Clearing denominators yields $$\left( f(m) + f(n) \right)^2 = 4 \cdot f(m) f(n)$$so $$0 = f(m)^2 - 2 \cdot f(m) f(n) + f(n)^2 = \left(f(m) - f(n) \right)^2$$which implies $f(m) = f(n)$, i.e. $m = n$. $\blacksquare$