For part A, I've tried solving each equation for $x$, then replacing $x^2$ in one equation with the squared results of the other, but have reached a point from which I have no idea how to proceed.
Since I kind of need the results of part A (or at least part B) to solve the rest of the questions, I'm just a bit stuck.
Note that I've also tried solving for $x$ and setting the equations equal to each other, as well as solving for $x^2$ and doing the same. They lead back to the same result.
For part B, I... I don't even know what's going on with $y=\frac{2}{3}$.
The Questions
a. By hand, show that $x^2+y^2+z^2=1$ and $3x + 9yz = 5$ intersect at precisely two points.
b. Find these two points of intersection, exactly.
c. Show that the two surfaces have common tangent planes at those respective points.
d. Find the equations of the two common tangent planes.
e. Use Mathematica to give a "good" graph of the two surfaces, along with the tangent planes.
My Work
Part A
$x^2 = 1-y^2-z^2$
$3x = 5-9yz$
$x = \frac{5}{3} - 3yz$
$x^2 = \frac{25}{9} - 10yz + 9y^2z^2$
$\frac{25}{9}-10yz+9y^2z^2=1-y^2-z^2$
$\frac{16}{9} = 10yz-9y^2z^2-y^2-z^2$
Part B
$y = \frac{2}{3}$ ---> $x = \frac{5}{3}-3(\frac{2}{3})z = \frac{5}{3} - 2z$
$\frac{25}{9}+\frac{4}{9}-\frac{20z}{3}-4z^2+z^2=1$
$\frac{29}{9}-\frac{20z}{3}-3z^2=1$
$\frac{29-60z-27z^2}{9}=1$
$29-60z-27z^2=9$
$20-60z-27z^2=0$
$z^2-\frac{20z}{9}-\frac{20}{27}+\frac{20}{27}=0+\frac{20}{27}$
$z^2-\frac{20z}{3}+(\frac{10}{9})^2=(z+\frac{10}{9})^2=\frac{160}{81}$
$z + \frac{10}{9} = \sqrt{\frac{160}{81}} = \frac{-10±4\sqrt{10}}{9}$
$z = \frac{-10+4\sqrt{10}}{9}$ ---> $x = \frac{35-8\sqrt{10}}{9}$
$z = \frac{-10-4\sqrt{10}}{9}$ ---> $x = \frac{35+8\sqrt{10}}{9}$
.
$y = \frac{-2}{3}$ ---> $x = \frac{5}{3}+2z$
$\frac{25}{9}+\frac{4}{9}+\frac{20z}{3}+4z^2+z^2=1$
$\frac{29}{9}+\frac{20z}{3}+5z^2=1$
$\frac{29+60z+45z^2}{9}=1$
$29+60z+45z^2=9$
$20+60z+45z^2=0$
$\frac{45z^2}{45}+\frac{60z}{45}+\frac{20}{45}=z^2+\frac{4z}{3}+\frac{4}{9}=0$
$(z=\frac{2}{3})^2=0$
$z=\frac{-2}{3}$ ---> $x=\frac{1}{3}$
A bit more simplification will give you $$(81y^2+9)z^2-90yz+(9y^2+16)=0\ .$$ Treat this as a quadratic in $z$: for real solutions, the discriminant must be non-negative, $$(90y)^2-4(81y^2+9)(9y^2+16)\ge0\ .$$ A bit of work turns this into $$(9y^2-4)^2\le0$$ and so $y=\pm\frac23$.
Since this was basically what you were asking I'll leave the rest up to you :)