Answer
$${\mathbf{G}(x, y)=2 x y e^{x^2}\mathbf{i}+ e^{x^2}\mathbf{j}+ \mathbf{k}}$$
Work Step by Step
Given
$$g(x, y,z)=z+y e^{x^2}$$
Since \begin{align}\mathbf{G}(x,y)&=M \mathbf{i}+N\mathbf{j}+P\mathbf{k}\\
&=g_{x}(x, y)\mathbf{i}+g_{y}(x, y)\mathbf{j}+g_{z}(x, y)\mathbf{k}
\end{align}
As, we have
\begin{array}{l}
{g_{x}(x, y)=\frac{\partial g(x,y)}{\partial x}=2xy e^{x^2} } \\
{g_{y}(x, y)=\frac{\partial g(x,y)}{\partial y}= e^{x^2} } \\
{g_{z}(x, y)=\frac{\partial g(x,y)}{\partial z}=1 } \\ \end{array}
So, we get
$${\mathbf{G}(x, y)=2 x y e^{x^2}\mathbf{i}+ e^{x^2}\mathbf{j}+ \mathbf{k}}$$
