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 $c(x,y), c(y, z), c(x,z)$?
On
An alternative approach: Define $f(x,y,z)=\int_\Gamma \bar v\cdot ds$ where $\Gamma$ is a smooth curve joining the origin to the point $(x,y,z)$. Since you know that this line integral is path independent, you can choose any convenient curve such as the line segment joining these two points. This line segment is conveniently parameterized as $(tx,ty,tz)$, $0\le t\le 1$, giving $$f(x,y,z) = \int_0^1\bar v(tx,ty,tz)\cdot(x,y,z)\,dt = \int_0^1 2x^2t+2y^2z^3t^4+3y^2z^3t^4\,dt = \frac12x^2+y^2z^3.$$ Choosing a different starting point for $\Gamma$ amounts to changing the constant of integration.
$$ f_x = x \implies f = \frac{1}{2}x^2 + g(y,z) $$ 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 $$ f = \frac{1}{2}x^2 + y^2z^3 + c $$