If $f\left(x\right)+g\left(x\right)=e^{-x}$ where $f\left(x\right)$ is an even function and $g\left(x\right)$ is odd function, then $f\left(x\right)=$

Given $f\left(x\right)+g\left(x\right)=e^{-x}$            ..... (1)

Given that $f\left(x\right)$ is even function, so that $f\left(-x\right)=f\left(x\right)$ and $g\left(x\right)$ is odd function, so that $g\left(-x\right)=-g\left(x\right)$

Replace $x$ by $-x$ in the given functional equation 

Hence, 

 $f\left(-x\right)+g\left(-x\right)=e^x\Rightarrow f\left(x\right)-g\left(x\right)=e^x$       ...... (2)

By adding both (1) and (2)

We get 

$2f\left(x\right)=e^x+e^{-x}$

Therefore, $f\left(x\right)=\boxed{\frac{e^x+e^{-x}}{2}}$