Relation between field and potential

$\vec E = -\left(\frac{\partial V}{\partial x}\hat i + \frac{\partial V}{\partial y}\hat j + \frac{\partial V}{\partial z}\hat k\right)$
$\vec E = -(6y)\hat i - (6x - 1 + 2z)\hat j - (2y)\hat k$
At (1, 1, 0): $\vec E = -(6\hat i + 5\hat j + 2\hat k)$

$\vec E = -\frac{\partial V}{\partial x}\hat i - \frac{\partial V}{\partial y}\hat j - \frac{\partial V}{\partial z}\hat k = -[(6 - 8y)\hat i + (-8x - 8 + 6z)\hat j + (6y)\hat k]$
At (1, 1, 1): $\vec E = 2\hat i + 10\hat j - 6\hat k$
$|\vec E| = \sqrt{2^2 + 10^2 + 6^2} = \sqrt{140} = 2\sqrt{35}$ N/C
$F = qE = 2\times 2\sqrt{35} = 4\sqrt{35}$ N

The electric field is the negative gradient of the potential: $E_x = -\frac{\partial V}{\partial x}$, $E_y = -\frac{\partial V}{\partial y}$, $E_z = -\frac{\partial V}{\partial z}$
$V = 4x^2$ depends only on x, so $E_y = E_z = 0$.
$E_x = -\frac{\partial}{\partial x}(4x^2) = -8x$
At the point (1, 0, 2), x = 1: $E_x = -8$ V/m
The negative sign shows the field is along the negative X-axis.
So the field is 8 V/m along the negative X-axis.