Answer
$$ \bar{x}=-1\ \ \text{ and }\ \ \bar{y}=\frac{1}{4}$$
Work Step by Step
Since
\begin{align*}
M&=\int_{-\infty}^{0} \int_{0}^{e^{x}} d y d x\\
&=\int_{-\infty}^{0} e^{x} d x=\lim _{b \rightarrow-\infty} \int_{b}^{0} e^{x} d x\\
&=1-\lim _{b \rightarrow-\infty} e^{b}=1\\
M_{y}&=\int_{-\infty}^{0} \int_{0}^{e^{x}} x d y d x\\
&=\int_{-\infty}^{0} x e^{x} d x\\
&=\lim _{b \rightarrow-\infty} \int_{b}^{0} x e^{x} d x\\
&=\lim _{b \rightarrow-\infty}\left[x e^{x}-e^{x}\right]_{b}^{0}\\
&=-1-\lim _{b \rightarrow-\infty}\left(b e^{b}-e^{b}\right)\\
&=-1\\
M_{x}&=\int_{-\infty}^{0} \int_{0}^{e^{x}} y d y d x\\
&=\frac{1}{2} \int_{-\infty}^{0} e^{2 x} d x\\
&=\frac{1}{2} \lim _{b \rightarrow-\infty} \int_{b}^{0} e^{2 x} d x\\
&=\frac{1}{4}
\end{align*}
Then
$$ \bar{x}=-1\ \ \text{ and }\ \ \bar{y}=\frac{1}{4}$$
