Find the the equation of the plane that tangent to $x^2+2y^2+4z^2+xy+3yz=1$ and is paralel to $y=0$ plane
first I found the gradient vector $\nabla f\left( x,y,z\right)=(2x+y)i+(4y+x+3z)j+(8z+3y)k $
and I equalized this to normal vector which is $(0,1,0)$
$2x_0+y_0=0$
$4y_0+x_0+3z_0=1$
$8z_0+3y=0$
and I found $y=\frac{8}{19}$
so the plane's equation should be $0(x-x_0)+(y-\frac{8}{19})+0(z-z_0)=0$
Is this correct?
You are correct that the tangent plane will have an equation of the form $y=y_0$, but you should have set the gradient equal to $(0,\lambda,0)$, since we do not know a priori that the gradient has unit magnitude. Then we can solve for $(x_0,y_0,z_0)$ in terms of $\lambda$. When we plug $(x_0,y_0,z_0)$ back into our equation for the surface, we can solve for $\lambda$. We are not be surprised to obtain 2 solutions for $\lambda$. That just means there are 2 tangent planes of the desired form. We can use the values of $\lambda$ to obtain the tangent points $(x_0,y_0,z_0)$ and the equations of the tangent planes.