Answer
$${\mathbf{G}(x, y)=3\cos (3x)\cos (4y) \mathbf{i}-4\sin (3x)\sin (4y) \mathbf{j}}$$
Work Step by Step
Given
$$g(x, y)=\sin (3x)\cos (4y)$$
Since \begin{align}\mathbf{G}(x,y)&=M \mathbf{i}+N\mathbf{j}\\
&=g_{x}(x, y)\mathbf{i}+g_{y}(x, y)\mathbf{j}
\end{align}
As, we have
\begin{array}{l}
{g_{x}(x, y)=\frac{\partial g(x,y)}{\partial x}=3\cos (3x)\cos (4y)} \\
{g_{y}(x, y)=\frac{\partial g(x,y)}{\partial y}=-4\sin (3x)\sin (4y)} \\ \end{array}
So, we get
$${\mathbf{G}(x, y)=3\cos (3x)\cos (4y) \mathbf{i}-4\sin (3x)\sin (4y) \mathbf{j}}$$
