The conjugate function of the $f(x) = e^x$.
The conjugate function is defined as $f^*(y) = \sup_{x \in \operatorname{dom}(f)}\{x^Ty-f(x)\}$. In this case we should find the supremum of the $xy-e^x$ over the domain of $f(x)$ that is over $\mathbb{R}$. Now $e^x$ grows faster than $yx$ so if $y > 0$ then $xy-e^x$ will go to the $-\infty$ as $x \to \infty$. At the same time if $y < 0$ we will have as $x \to \infty$ that $xy-e^x$ goes to $-\infty$. If $y > 0$ and $x \to -\infty$ we will have $-\infty$ since $e^x \to 0$ in that case. If $y < 0$ and $x \to -\infty$ we will have $\infty$ for the same reason. If $y = 0$ we will have $\infty$.
Best Answer
IA AM ASSUMING that $f^*(y):=\sup\{xy-f(x):\, x\in {\rm Dom}\,(f)\}$.
For $y>0$, take $g_y(x)=xy-e^x$. Then $g_y'(x)=y-e^x=0$ $\iff$ $x=\ln y$ and it is a MAXIMUM ($g_y'(x)>0$ if $x<\ln y$ and $g_y'(x)<0$ is $x>\ln y$).
So, for $y>0$, one has that $$f^*(y)=g_y(\ln y)=y\ln y-e^{\ln y}=y(\ln y-1).$$
If $y=0$, it is clear that $f^*(y)=+\infty$ (it does not exist) since $\lim_{x\to+\infty}e^x=+\infty$.
If $y< 0$, $xy-e^x\le0$ and $\lim_{x\to-\infty}(xy-e^x)=+\infty-e^{-\infty}=+\infty$, so $f^*(y)=+\infty$ (so, it does not exists).