Answer
${\mathbf{F}(x, y)=\frac{x}{\sqrt{x^2+4y^2+z^2}} \mathbf{i}+\frac{4y}{\sqrt{x^2+4y^2+z^2}}\mathbf{j}+\frac{z}{\sqrt{x^2+4y^2+z^2}} \mathbf{k}}$
Work Step by Step
Given
$$f(x, y,z)=\sqrt{x^2+4y^2+z^2}=(x^2+4y^2+z^2)^\frac{1}{2}$$
Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}+P\mathbf{k}\\
&=f_{x}(x, y)\mathbf{i}+f_{y}(x, y)\mathbf{j}+f_{z}(x, y)\mathbf{k}
\end{align}
As, we have
\begin{array}{l}
{f_{x}(x, y)=\frac{\partial f(x,y)}{\partial x}=\frac{1}{2}(2x)(x^2+4y^2+z^2)^{-\frac{1}{2}}=\frac{x}{\sqrt{x^2+4y^2+z^2}}} \\
{f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=\frac{1}{2}(8y)(x^2+4y^2+z^2)^{-\frac{1}{2}}=\frac{4 y}{\sqrt{x^2+4y^2+z^2}}} \\
{f_{z}(x, y)=\frac{\partial f(x,y)}{\partial z}=\frac{1}{2}(2z)(x^2+4y^2+z^2)^{-\frac{1}{2}}=\frac{z}{\sqrt{x^2+4y^2+z^2}}} \\ \end{array}
So, we get
$${\mathbf{F}(x, y)=\frac{x}{\sqrt{x^2+4y^2+z^2}} \mathbf{i}+\frac{4y}{\sqrt{x^2+4y^2+z^2}}\mathbf{j}+\frac{z}{\sqrt{x^2+4y^2+z^2}} \mathbf{k}}$$
