Sum of inverses is inverse of sum:
$$(f(x)+g(x))^{-1} = f^{-1}(x) + g^{-1}(x)$$
It is not true for all functions.
For example $x$ is its own function inverse, but $x+x = 2x$ but the inverse is $x/2$ not $2x$.
Which functions make it true?
$$(f(x)+g(x))^{-1} = f^{-1}(x) + g^{-1}(x)$$
By $^{-1}$ I mean inverse of the function, not $1$ divided by it.
If you don't mind complex solutions, you can do it: for any $b \ne 0$ take $$ f(x) = b x,\ g(x) = b \omega x$$ Thus $$ f^{-1}(x) = x/b, \ g^{-1}(x) = x/(b \omega)$$ and $$ f(x) + g(x) = b (1+\omega) x, \ (f+g)^{-1}(x) = x/(b (1+\omega))$$ So $(f+g)^{-1} = f^{-1} + g^{-1}$ if $$ \frac{1}{1+\omega} = 1 + \frac{1}{\omega} $$ which is true if $1 + \omega + \omega^2 = 0$. The roots of this are $\omega = \frac{1 \pm \sqrt{3} i}{2}$, the primitive cube roots of $-1$.