The task is: "Find the point(s) on $x^2+y^2+z^2 = 8$, where the tangent plane is parallel to the plane $x-y+2z=0$"
What I did is: $f(x,y,z) = x^2+y^2+z^2-8$ grad$f = (2x,2y,2z)$
normal vector of $x-y+2z=0$ is $n=(1,-1,2)$
grad$f \parallel n \iff 2x=1, 2y=-1, 2z=2$
Therefore wanted point is $(\frac{1}{2},-\frac{1}{2};1)$ Is this ok?
You're on the right track.
Note that for the gradient of $f$ to be parallel to the normal vector, the vectors don't have to be equal but scalar multiples; so you get: $$\mbox{grad}\,f \parallel n \iff (2x,2y,2z)=k\cdot(1,-1,2)$$ Substitution of $x$, $y$ and $z$ as a function of $k$ into the equation of the sphere gives you a quadratic equation in $k$; the solutions correspond to points on the sphere with the desired gradient.