Answer
$${\mathbf{F}(x, y)=2 x \mathbf{i}+4 y \ \mathbf{j}}$$
Work Step by Step
Given $$f(x, y)=x^{2}+2 y^{2}$$ Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}\\ &=f_{x}(x, y)\mathbf{i}+f_{y}(x, y)\mathbf{j} \end{align} As, we have \begin{array}{l} {f_{x}(x, y)=\frac{\partial f(x,y)}{\partial x}=2 x} \\ {f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=4 y} \\ \end{array} So, we get $${\mathbf{F}(x, y)=2 x \mathbf{i}+4 y \mathbf{j}}$$
