please provide answer for this question Show that the vector field by ⃗= (x^2+4yz)̂+ (y^2+4zx)̂+...

We are saked to show that the vector field
$\overrightarrow{f}=(x^2+4yz)\overrightarrow{i}+(y^2+4zx)\overrightarrow{j}+(z^2+4xy)\overrightarrow{k}$
is irrational. Also we are asked to find scalar potential. Since we know that a vector filed $\overrightarrow{A}$ is irrational if
$curl\overrightarrow{A}=\overrightarrow{0}$ or $\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{A}=\overrightarrow{0}$
Also with respect to a irrational field $\overrightarrow{A}$ there exist a scalar potential $\oslash$ such that $\overrightarrow{A}=\overrightarrow{\mathrm{\nabla }}\oslash$
Now solving for proving $\overrightarrow{f}$ irrational we will find $curl\overrightarrow{f}$
$curl\overrightarrow{f}=\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{f}=\left[\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ f_x& f_y& f_z\end{array}\right]$
Where $\frac{\partial }{\partial x},\frac{\partial }{\partial y}$ and $\frac{\partial }{\partial z}$ are partial derivatives with respect to and $\overrightarrow{f}=f_x\overrightarrow{i}+f_y\overrightarrow{i}+f_z\overrightarrow{k}$ x,y and z respectively.
Now $\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{f}=\left[\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x^2+4yz& y^2+4zx& z^2+4xy\end{array}\right]$
$=\overrightarrow{i}[\frac{\partial }{\partial y}(z^2+4xy)-\frac{\partial }{\partial z}(y^2+4zx)]-\overrightarrow{j}[\frac{\partial }{\partial x}(z^2+4xy)-\frac{\partial }{\partial z}(x^2+4yz)]+\overrightarrow{k}[\frac{\partial }{\partial x}(y^2+4zx)-\frac{\partial }{\partial y}(x^2+4yz)]$
$=\overrightarrow{i}(4x-4x)-\overrightarrow{j}(4y-4y)+\overrightarrow{k}(4z-4z)$
$=\overrightarrow{i}(0)-\overrightarrow{j}(0)+\overrightarrow{k}(0)$
$\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{f}=\overrightarrow{0}$
Since $\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{f}=\overrightarrow{0}$
so $\overrightarrow{f}$ is irrational vector field.
Now for scalar potential $\oslash$
since $\overrightarrow{f}=\mathrm{\nabla }\oslash =(\overrightarrow{i}\frac{\partial }{\partial x}+\overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z})\oslash$
$\overrightarrow{f}=\frac{\partial \oslash }{\partial x}\overrightarrow{i}+\frac{\partial \oslash }{\partial y}\overrightarrow{j}+\frac{\partial \oslash }{\partial z}\overrightarrow{k}$