<s>[INST] Given a vector field $\mathbf{H}$, find a vector field $\mathbf{F}$ and a scalar field g, such that $\mathbf{H}$ = curl(F) + ∇(g). Let$\;\mathbf{H}(x,y,z) = x^2y\mathbf{i}+y^2z\mathbf{j}+z^2x\mathbf{k}$. Find a vector field $\mathbf{F}$ and a scalar field g, such that $\mathbf{H}$ = curl(F) + ∇(g).
I took divergence on both sides which gave me $2xy+2yz+2xz=∇^2g$. I took curl on both sides which gave me three horrible equations one of which I have written down below: 
$$\frac{\partial^2f_1}{\partial y^2}-\frac{\partial^2f_2}{\partial x\partial y}-\frac{\partial^2f_3}{\partial x\partial z}+\frac{\partial^2f_1}{\partial z\partial y}=-y^2$$
where I have taken
$$\mathbf{F}=f_1\mathbf{i}+f_2\mathbf{j}+f_3\mathbf{k}$$
I tried to assign some values hoping to work out the rest but this gave me nothing. Please help. 
 [/INST] Below a slightly different approach which first gives $\mathbf{F}$
and then $G$.
The problem at hand is quite similar to that of finding the vector and
scalar potentials in terms of the electric and magnetic fields in
electrodynamics. In that case there are standard solutions. But here, since $%
\mathbf{H}$ is not decaying for large $\mathbf{x}$, they do not apply. But
it seems that a suitable $\mathbf{F}$ can easily be found. Thus we have$$
\mathbf{H}=\left(
\begin{array}{c}
x_{1}^{2}x_{2} \\
x_{2}^{2}x_{3} \\
x_{3}^{2}x_{1}
\end{array}
\right) =\partial _{\mathbf{x}}\times \mathbf{F}+\partial _{\mathbf{x}}G..
$$
We can assume
$
\partial _{\mathbf{x}}\cdot \mathbf{F}=0.
$
Now
\begin{eqnarray*}
\partial _{\mathbf{x}}\times \mathbf{H} &=&\left(
\begin{array}{c}
-x_{2}^{2} \\
-x_{3}^{2} \\
-x_{1}^{2}
\end{array}
\right) =\partial _{\mathbf{x}}\times (\partial _{\mathbf{x}}\times \mathbf{F%
})=\mathbf{-\partial }_{\mathbf{x}}^{2}\mathbf{F}, \\
\partial _{\mathbf{x}}\cdot \mathbf{H} &\mathbf{=}%
&2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})=\mathbf{-\partial }_{\mathbf{x}}^{2}G.
\end{eqnarray*}
Thus,  $\mathbf{-\partial }_{\mathbf{x}}^{2}F_{1}=-x_{2}^{2}$ and assuming $
F_{1}=F_{1}(x_{2})$ we find $F_{1}(x_{2})=a+bx_{2}+cx_{2}^{2}+dx_{2}^{3}+
\frac{1}{12}x_{2}^{4}$. Setting the coefficients $a,b,c,d$ equal to $0$, we
obtain
$$
\mathbf{F}(\mathbf{x})=\frac{1}{12}\left(
\begin{array}{c}
x_{2}^{4} \\
x_{3}^{4} \\
x_{1}^{4}
\end{array}
\right) ,
$$
which is divergence-free. Now
$$
\partial _{\mathbf{x}}\times \mathbf{F}=\left(
\begin{array}{c}
-\partial _{3}F_{2} \\
-\partial _{1}F_{3} \\
-\partial _{2}F_{1}
\end{array}
\right) =-\frac{1}{3}\left(
\begin{array}{c}
x_{3}^{3} \\
x_{1}^{3} \\
x_{2}^{3}
\end{array}
\right) ,
$$
and
$$
\mathbf{H}-\partial _{\mathbf{x}}\times \mathbf{F}=\left(
\begin{array}{c}
x_{1}^{2}x_{2} \\
x_{2}^{2}x_{3} \\
x_{3}^{2}x_{1}
\end{array}
\right) +\frac{1}{3}\left(
\begin{array}{c}
x_{3}^{3} \\
x_{1}^{3} \\
x_{2}^{3}
\end{array}
\right) =\left(
\begin{array}{c}
x_{1}^{2}x_{2}+\frac{1}{3}x_{3}^{3} \\
x_{2}^{2}x_{3}+\frac{1}{3}x_{1}^{3} \\
x_{3}^{2}x_{1}+\frac{1}{3}x_{2}^{3}
\end{array}
\right) =\partial _{\mathbf{x}}G.
$$
\begin{eqnarray*}
\partial _{1}G &=&x_{1}^{2}x_{2}+\frac{1}{3}x_{3}^{3}\Longrightarrow G=\frac{
1}{3}x_{1}^{3}x_{2}+\frac{1}{3}x_{3}^{3}x_{1}+M \\
\partial _{2}G &=&x_{2}^{2}x_{3}+\frac{1}{3}x_{1}^{3}=\frac{1}{3}
x_{1}^{3}+\partial _{2}M \\
\partial _{2}M &=&x_{2}^{2}x_{3}\Rightarrow M=\frac{1}{3}x_{2}^{3}x_{3}+N
\end{eqnarray*}
Setting $N=0$,
$$
G=\frac{1}{3}(x_{1}^{3}x_{2}+x_{3}^{3}x_{1}+x_{2}^{3}x_{3})$$
</s>