First, write F=4x^2-3y+z^2=0 as whole equation for the surface (level surface format), then we partially differentiate each term:
DF/Dx=8x; DF/Dy=-3; DF/Dz=2z where D is used to designate partial differential.
Plugging in (0,1,sqrt(3)) we get: DF/Dx=0, DF/Dy=-3; DF/Dz=2sqrt(3).
The normal is vector N=(0,-3,2sqrt(3)) but we need the equation of the tangential plane and for that we need vector r=(x,y,z) and the vector dot product (scalar product) of r.N=(0,1,sqrt(3)).N.
This gives us 0-3y+2zsqrt(3)=-3+6=3. The equation of the plane is 2zsqrt(3)-3y=3.