<s>[INST] Exhibiting $f$ such that a vector field $\overline{v} = \nabla f$? The first is suppose $\overline{v} = (x, 2yz^3, 3y^2z^2)$ and we want to exhibit an $f$ such that $\overline{v} = \nabla f$.  Since $\nabla \times \overline{v}=0$, $f$ exists and we can integrate: 
$$\displaystyle  \int f \,dx = \frac{1}{2}x^2+c(y,z), ~ \int f \,dy = y^2z^3+c(x,z), ~ \int f \,dz = y^2z^3+c(y,z)$$
So $\displaystyle f = \frac{1}{2} x+c(y, z) = y^2z^3+c(x,z) = y^2z^3+c(x,z)$. How do you determine $c(x,y), c(y, z), c(x,z)$?
 [/INST] $$ f_x = x \implies f = \frac{1}{2}x^2 + g(y,z) $$
We know $f_y = 2yz^3$ so
$$ f_y = 0 + g_y = 2yz^3 \implies g(y, z) = y^2z^3 + h(z) $$
So now we have
$$ f = \frac{1}{2}x^2 + y^2z^3 + h(z) $$
We know $f_z = 3y^2z^2$ so
$$ f_z = 0 + 3yz^2 + h'(z) = 3y^2z^2 \implies h(z) =c, c\in\mathbb{R} $$
And we have that 
$$ f = \frac{1}{2}x^2 + y^2z^3 + c $$
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