Answer
\begin{align}
d z=\left(e^{x^{2}+y^{2}}+e^{-x^{2}-y^{2}}\right)(x d x+y d y)
\end{align}
Work Step by Step
Given $$ z=\frac{1}{2}\left(e^{x^{2}+y^{2}}-e^{-x^{2}-y^{2}}\right)$$
Since $$dz=\frac{\partial z}{\partial x} dx+\frac{\partial z}{\partial y} dy ,$$
$$\frac{\partial z}{\partial x} =\frac{1}{2}\left((2x )e^{x^{2}+y^{2}}-(-2x)e^{-x^{2}-y^{2}}\right)=x\left(e^{x^{2}+y^{2}}+e^{-x^{2}-y^{2}}\right) ,$$
and
$$\frac{\partial z}{\partial y} =\frac{1}{2}\left((2y )e^{x^{2}+y^{2}}-(-2y)e^{-x^{2}-y^{2}}\right)=y\left(e^{x^{2}+y^{2}}+e^{-x^{2}-y^{2}}\right)$$
then we get
\begin{align}
d z&=x\left(e^{x^{2}+y^{2}}+e^{-x^{2}-y^{2}}\right) d x+y\left(e^{x^{2}+y^{2}}+e^{-x^{2}-y^{2}}\right) dy \\ &=\left(e^{x^{2}+y^{2}}+e^{-x^{2}-y^{2}}\right)(x d x+y d y)
\end{align}
