Answer
$$\operatorname{proj}_{{g}} {f} =\frac{2(e^2-2e)}{e^2-1}e^{-x}.$$
Work Step by Step
Let $f(x)=x, \quad g(x)=e^{-x}, \quad C[0,1], \quad\langle f, g\rangle=\int_{0}^1 f(x)g(x) d x.$
We have $$\langle f, g\rangle=\int_{0}^1 xe^{-x} d x=\left[-xe^{-x}-e^{-x}\right]_{0}^1=1-\frac{2}{e}$$
$$\langle g, g\rangle=\int_{0}^{1}e^{-2x}d x= -\frac{1}{2}\left[e^{-2x}\right]_{0}^{1}=\frac{1}{2}(1-\frac{1}{e^2}),$$
Now, the orthogonal projection of $f$ onto $g$ is given by $$\operatorname{proj}_{{g}} {f} =\frac{\langle{f}, {g}\rangle}{\langle{g}, {g}\rangle} {g}=\frac{1-\frac{2}{e}}{\frac{1}{2}(1-\frac{1}{e^2})}e^{-x}=\frac{2(e^2-2e)}{e^2-1}e^{-x}.$$
