Answer
Answer (a): $\vec{F}$ is conservative for all $x$, $y$.
Answer(b): $\vec{F}$ is conservative for all $x$, $y$ and $z$
Work Step by Step
Answer of part (a)
\begin{equation}
\phi = 2 y ^ { 2 } + 3 x ^ { 2 } y - x y ^ { 3 }
\end{equation}
\begin{equation}
\nabla \phi = \phi _ {x } i + \phi _ { y } j = \left( 6 x y - y ^ { 3 } \right)i + \left( 4 y + 3 x ^ { 2 } - 3 x y ^ { 2 } \right) j = F
\end{equation}
\begin{equation}
where:
\quad \phi _ { x } = \frac { \partial \phi } { \partial x }, \quad \phi _ { y } = \frac { \partial \phi } { \partial y }
\end{equation}
Therefore, $\vec{F}$ is conservative for all $x$, $y$.
Answer of part (b)
\begin{equation}
\phi = x \sin z + y \sin x + z \sin y
\end{equation}
\begin{equation}
\nabla \phi = \phi _ { x } i + \phi _ { y } j + \phi _ { z } k
\end{equation}
\begin{equation}
\phi _ {x } = \frac { \partial \phi } { \partial x} = \sin z + y \cos x
\end{equation}
\begin{equation}
\phi _ { y } = \frac { \partial \phi } { \partial y } = \sin x + 2 \cos y
\end{equation}
\begin{equation}
\phi _ { z } = \frac { \partial \phi } { \partial z } = x \cos z + \sin y
\end{equation}
$\vec{\nabla}\phi = (\sin(z) + y\cos(x))\mathbf{i} + (\sin(x) + z\cos(y))\mathbf{j} + (x\cos(z) + \sin(y))\mathbf{k} - \mathbf{F}$
Therefore, $\vec{F}$ is conservative for all $x$, $y$ and $z$
