Tangent planes to $2+x^2+y^2$ and that contains the $x$ axis

So far you have worked out that the tangent plane to the surface at $(x_0,y_0,x_0^2 + y_0^2+2)$ has equation $$ 2xx_0 + 2yy_0-z-x_0^2-y_0^2+2=0 \tag{1} $$

If this plane contains the $x$-axis, it contains all points $(x,y,z)$ with $y=z=0$. So the equation reduces to $$ 2x x_0 = x_0^2 + y_0^2 -2 \tag{2} $$ Again, we are looking for pairs $(x_0,y_0)$ such that (2) is true for all $x$. This is different from looking for triples $(x,x_0,y_0)$ satisfying (2). If we substitute $x=\frac{x_0}{2}$ into (2) we get $$ x_0^2 = x_0^2 + y_0^2 - 2 \implies y_0^2 = 2 \implies y_0 = \pm\sqrt{2} $$ Now equation (2) reduces to $2x x_0 = x_0^2$. If we substitute $x=0$ we get $$ x_0^2 = 0 \implies x_0=0 $$ Since the points are on the paraboloid, their $z$-coordinates satisfy $$ z_0 = x_0^2 + y_0^2 + 2 = (0)^2 + \left(\sqrt{2}\right)^2 +2 = 4 $$ So there are two points on the surface for which the tangent plane contains the $x$-axis: $(0,\sqrt{2},4)$ and $(0,-\sqrt{2},4)$. We can find the equations for the planes by substituting these points into (1). The first gives $$ 2x(0) + 2y(\sqrt{2}) - z - 0^2 - 2 + 2 = 0 \implies z = 2 \sqrt{2} y $$ The second is $z = -2\sqrt{2} y$.

Another way to think about this is to project into two dimensions. It's equivalent to asking: Which lines in the $yz$-plane are tangent to $z=y^2+2$ and pass through the origin? The line through $(y_0,y_0^2+2)$ and the origin has slope $\frac{y_0^2 + 2}{y_0}$. The line through $(y_0,y_0^2+2)$ tangent to $z=y^2+2$ has slope $2y_0$. So $$ \frac{y_0^2 + 2}{y_0} = 2y_0 \implies y_0^2 + 2 = 2y_0^2 \implies y_0^2 = 2 \implies y_0 = \pm\sqrt{2} $$