If potential (in volts) in a region is expressed as $V(x, y, z) = 6 xy - y + 2yz,$ the electric field (in N/C) at point $(1, 1, 0)$ is: 

Potential in a region 

$\mathrm{V}=6\mathrm{xy}-\mathrm{y}+2\mathrm{yz}$

As we know the relation between electric potential and electric field is $\overrightarrow{\mathrm{E}}=\frac{-\mathrm{dV}}{\mathrm{dx}}$

$\begin{array}{l}\overrightarrow{\mathrm{E}}=\left(\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{x}}\widehat{\mathrm{l}}+\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{y}}\widehat{\mathrm{j}}+\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{z}}\widehat{\mathrm{k}}\right)\\ \overrightarrow{\mathrm{E}}=[6\mathrm{y}\widehat{\mathrm{i}}+(6\mathrm{x}-1+2\mathrm{z})\widehat{ȷ}+(2\mathrm{y}\left)\widehat{\mathrm{k}}\right]\\ \overrightarrow{\mathrm{E}}(1,1,0)=-(6\widehat{\mathrm{i}}+5\mathrm{j}+2\widehat{\mathrm{k}})\end{array}$