A) The relation $x^2+z^2=1$ calls for the substitution $x=\cos t, z=\sin t$. Then $y=-x$ turns to $y=-\cos t$.
B) Indeed, $\cos^2t+(-\cos t)^2+2\sin^2t=2$. But you can establish this independently: $x^2+y^2+2z^2=x^2+(-x)^2+2z^2=2(x^2+z^2)=2$. The surface in question is an ellipsoid of revolution.
C) When projected onto $xz$, the curve is the unit circle. When projected onto $yz$, the curve is the unit circle. And when projected onto $xy$, the curve is a line segment with direction $-45°$. As is well known, the intersection of a cylinder and a plane is an ellipse (axes $1$ and $\sqrt2$).
You can check all of this by rotating space around the axis $z$ by $45°$, with the transform
$x=\frac1{\sqrt2}(u+v), y=\frac1{\sqrt2}(u-v), w=z$.
The cylinder and the plane become $\frac{(u+v)^2}2+w^2=1$ and $u=0$, so that $\frac{v^2}2+w^2=1$. This is the implicit equation of an ellipse centered at the origin, with axis $\sqrt2$ and $1$.
Also, $u=\frac{x+y}{\sqrt2}=0, v=\frac{x-y}{\sqrt2}=\sqrt2\cos t, w=\sin t$, which is the parametric equation of an ellipse.
The surface $S$ is defined as the set of all points in $\mathbb{R}^3$ whose coordinates satisfy $x^2 + y^2 = z^2$.
What is the image of your curve? Well, the points of the image of your curve are precisely points of the form $r(t) = (\sin t\cos t, \cos^2 t, \cos t)$. As $t$ ranges over all of $\mathbb{R}$, $r(t)$ ranges over all points in the image of your curve.
Thus, you want to ask: Do all points in the image of your curve have coordinates satisfying $x^2 + y^2 = z^2$? The important thing to note is that here $x,y,z$ just refer to the first, second, and third coordinates of your points. Well, the points of the image of your curve are of the points with coordinates $(\sin t\cos t,\cos^2 t,\cos t)$, and thus you want to check if the coordinates satisfy:
(first coordinate squared) + (second coordinate squared) = (third coordinate squared)
How do you check that? Well you simply verify that
$$(\sin t\cos t)^2 + (\cos^2 t)^2 = (\cos t)^2$$
for all $t\in\mathbb{R}$.
If this is true, then that's to say that for any $t\in\mathbb{R}$, the point $(\sin t\cos t,\cos^2 t,\cos t)$ satisfies (first coordinate squared) + (second coordinate squared) = (third coordinate squared), and thus lies on your surface $S$.
Since every point on your curve can be described in this way, you're done.
Best Answer
In this case since you have your parametrized curve given by $\gamma(t) = (t\cos{t}, t\sin{t}, t)$, one way of trying to find an algebraic equation satisfied by any point on your curve is writing $x = t\cos{t}$, $y = t\sin{t}$ and $z = t$. Now since you have sine and cosine, you have to think about the relations that these functions satisfy.
The easiest one, which you obviously know, is $\cos^2{t} + \sin^2{t} = 1$. Now in terms of $x, y, z$ you see that
$$\cos{t} = \frac{x}{z} \quad \text{and} \quad \sin{t} = \frac{y}{z}$$
Then this tells you that
$$\left( \frac{x}{z} \right )^2 + \left( \frac{y}{z} \right )^2 = 1$$
and this implies that $\ x^2 + y^2 = z^2 \ $ for any point $(x, y, z) = (t\cos{t}, t\sin{t}, t) \ $ on the curve $\gamma$.
So this tells you that the curve $\gamma$ is contained in the surface of equation $x^2 + y^2 = z^2$, which is the equation of a cone as already pointed out in the comments.
Finally since you needed to draw the curve in SAGE, the following is a plot of the curve and the cone. You can see how the curve wraps around the cone.
In case you want it, the SAGE code I used is the following.