$Part\ 1: \\[2ex]F=\left(x^2-yz\right)\hat{i}+\left(y^2-zx\right)\hat{j}+\left(z^2-xy\right)\hat{k}\ \\[2ex] \therefore{}r=xi+yj+zk \\[2ex] \displaystyle Curl\ F=\left\vert{}\begin{array}{ ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial{}}{\partial{}x} & \frac{\partial{}}{\partial{}y} & \frac{\partial{}}{\partial{}z} \\ x^2-yz & y^2-zx & z^2-xy \end{array}\right\vert{} \\[2ex]$

$\displaystyle =0-0+0=0 \\[2ex] \therefore{}F\ is\ a\ conservative\ field. \\[2ex] \therefore{}there\ exists\ a\ scalar\ potential\ of\ F\ such\ that\ \ F=\nabla{}\varphi{}\\[2ex] \displaystyle \therefore{}\left(x^2-yz\right)\hat{i}+\left(y^2-zx\right)\hat{j}+\left(z^2-xy\right)\hat{k}=i\frac{\partial{}\varphi{}}{\partial{}x}+j\frac{\partial{}\varphi{}}{\partial{}x}+\frac{\partial{}\varphi{}}{\partial{}x}\\[2ex]$

$\displaystyle Comparing\ both\ sides\ gives, \\[2ex] \displaystyle \frac{\partial{}\varphi{}}{\partial{}x}=x^2-yz;\frac{\partial{}\varphi{}}{\partial{}x}=y^2-zx;\frac{\partial{}\varphi{}}{\partial{}x}=z^2-xy \\[2ex]$

$\displaystyle Now\\[2ex] d\varphi{}=\frac{\partial{}\varphi{}}{\partial{}x}dx+\frac{\partial{}\varphi{}}{\partial{}y}dy+\frac{\partial{}\varphi{}}{\partial{}z}dz \\[2ex] \therefore{}\ d\varphi{}={(x}^2-yz)dx+{(y}^2-zx)dy+\left(z^2-xy\right)dz\\[2ex] \displaystyle on\ Integration, \\[2ex] \displaystyle \varphi{}=\frac{x^3}{3}-yz.x+\frac{y^3}{3}+\frac{z^3}{3}+C \\[2ex] \therefore{}Scalar\ potential \ of \ F=\varphi{}\\[2ex] \displaystyle =\frac{1}{3}\left(x^3+y^3+z^3\right)-xyz+C$