Answer
$\nabla $f transforms as a vector under rotations
Work Step by Step
$\bar y$ = +y cos $\phi$ + z sin $\phi$;
multiply by sin $\phi$:
y sin $\phi$ = +y sin$\phi$ cos $\phi$ + z $sin^2$ $\phi$.
z = −y sin $\phi$+ z cos $\phi$;
multiply by cos $\phi$:
z cos $\phi$ = −y sin $\phi$ cos $\phi$ + z $cos^2$ $\phi$
Add:
y sin $\phi$ + z cos $\phi$ = z($sin^2$ $\phi$ + $cos^2$ $\phi$) = z
Likewise, y cos $\phi$ − z sin $\phi$ = y.
So
$\frac{\partial y}{\partial\bar y}$ = cos $\phi$;
$\frac{\partial y}{\partial z}$ = -sin $\phi$;
$\frac{\partial z}{\partial \bar y}$ = sin $\phi$;
$\frac{\partial z}{\partial \bar z}$ = cos $\phi$;
Therefore
$\overline { (\nabla f)}_y$ = $\frac{\partial f}{\partial\bar y}$ = $\frac{\partial f}{\partial y}$ $\frac{\partial y}{\partial\bar y}$ + $\frac{\partial f}{\partial z}$ $\frac{\partial z}{\partial\bar y}$ = + cos $\phi$ $ (\nabla f)_y$ + sin $\phi$ $ (\nabla f)_z$
and
$\overline { (\nabla f)}_z$ = $\frac{\partial f}{\partial\bar z}$ = $\frac{\partial f}{\partial y}$ $\frac{\partial y}{\partial\bar z}$ + $\frac{\partial f}{\partial z}$ $\frac{\partial z}{\partial\bar z}$ = - sin $\phi$ $ (\nabla f)_y$ + cos $\phi$ $ (\nabla f)_z$
So $\nabla f$ transforms as a vector
