Answer
$f(x,y) =\dfrac{x^2}{y} -\dfrac{1}{y}+C$
Work Step by Step
The vector field $F$ is conservative, so we can write as: $F =\nabla f$
$M=\dfrac{2x}{y}\\ N=\dfrac{1-x^2}{y^2} \\ P=0$
$\dfrac{\partial N}{\partial z}=\dfrac{\partial P}{\partial y}=0\\ \dfrac{\partial M}{\partial z}=\dfrac{\partial P}{\partial x}=0 \\ \dfrac{\partial N}{\partial x}=\dfrac{\partial M}{\partial y}=\dfrac{-2x}{y^2}$
So, $ \dfrac{\partial f}{\partial x}=\dfrac{2x}{y} $
or,$f(x,y,z) =\dfrac{x^2}{y} +k (y) ...(a)$
and $\dfrac{\partial f}{\partial y}=\dfrac{1-x^2}{y}=\dfrac{-x^2}{y}+\dfrac{dk}{dy}$
or, $ \dfrac{ dk}{dy}=\dfrac{1}{y^2} \\k(y) =\dfrac{-1}{y} +C .....(b)$
Now, from equations (a) and (b), we have:
$f(x,y) =\dfrac{x^2}{y} -\dfrac{1}{y}+C$
