Proving that $\vec F=yz(2x+y+z)\hat i+zx(x+2y+z)\hat j+xy(x+y+2z)\hat k$ is conservative field I need to prove that $\vec F$ is conservative field

$$\vec F=yz(2x+y+z)\hat i+zx(x+2y+z)\hat j+xy(x+y+2z)\hat k$$

My attempt:
$\vec{F}$ is conservative iff $\nabla \times \vec{F} = 0$
$$
\begin{vmatrix}
&\hat i&\hat j &\hat k\\
&\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\
&yz(2x+y+z)&zx(x+2y+z)&xy(x+y+2z)
\end{vmatrix}
$$
$\color{blue}\Longrightarrow$
$$
=\begin{vmatrix}
&\hat i&\hat j &\hat k\\
&2yz&2xz&2xy\\
&yz(2x+y+z)&zx(x+2y+z)&xy(x+y+2z)
\end{vmatrix}
$$
$$=[x+y+2z-2]\hat i -[x+y+2z-2]\hat j +[x+2y+z-2]\hat k\color{red}{\neq 0}$$

Where am I wrong?

 A: Note that you have a 'potential'
$$
\phi = x^2yz + xy^2z + xyz^2
$$
so that
$$
\vec{F} = \vec\nabla \phi,
$$
whence
$$
\vec\nabla \times \vec{F} = \vec\nabla \times \vec\nabla \phi = 0,
$$
so it is conservative.
A: It's just as easy to use the definition:
$\vec F$ is conservative $\Leftrightarrow $ there is a $\varphi :\mathbb R^{3}\to \mathbb R$ such that $\nabla \varphi =\vec F$. The method for finding $\varphi $ is straightforward and has the advantage that it is consistent; it will tell you if such a $\varphi $ exists. 
If there is a $\varphi $ with the advertised property, then
$\frac{\partial \varphi }{\partial x}=yz(2x+y+z)$ which implies that 
$\varphi =yz(x^{2}+yx+zx)+g(y,z)$. Then,
$\frac{\partial \varphi }{\partial y}=zx^{2}+2zxy+z^{2}x+\frac{\partial g(y,z)}{\partial y}$ and this must be equal to 
$zx(x+2y+z)$ so that 
$\frac{\partial g(y,z)}{\partial y}=0$. This means that 
$g(y,z)=f(z)+c$ for some constant $c\in \mathbb R$. So we have, up to now
$\varphi =yz(x^{2}+yx+zx)+f(z)+c$ from which 
$\frac{\partial g}{\partial z}=yx^{2}+y^{2}x+2zyx+f'(z)$ and this must be equal to 
$xy(x+y+2z)$ so that 
$f'(z)=0\Rightarrow f(z)=c_{1}$. Then, finally we have
$\varphi =yz(x^{2}+yx+zx)+C$ where $C=c+c_{1}$.
