Answer
$E = (-12~N/C)~\hat{i}+ (12~N/C)~\hat{j}$
Work Step by Step
We can find $E_x$:
$E_x = -\frac{dV}{dx}$
$E_x = (-4.0~x)~N/C$
$E_x = (-4.0)~(3.0)~N/C$
$E_x = -12~N/C$
We can find $E_y$:
$E_y = -\frac{dV}{dy}$
$E_y = (6.0~y)~N/C$
$E_y = (6.0)~(2.0)~N/C$
$E_y = 12~N/C$
We can express the electric field in unit-vector notation:
$E = (-12~N/C)~\hat{i}+ (12~N/C)~\hat{j}$
