Find all points on the surface where the normal line passes through the origin
Since the normal vector passes through $(0,0,0)$ and $(x, y, z)$ then
$( 4 x^3, 4 y^3 , 4 z^3 ) = \alpha (x, y, z) $ for some $\alpha$
Hence, $ \alpha = 4 x^2 = 4 y^2 = 4 z^2 $
From which $|x| = |y| = |z| $
The solution of which is $(x, y, z) = \beta (\pm 1, \pm 1, \pm 1) $
Plugging this into $x^4 + y^4 + z^4 = 3 $ we deduce that $\beta = \pm 1$
Hence there are $8$ solutions for $(x,y,z)$ and they are $(\pm 1, \pm 1, \pm 1)$