The two equations are -
$x^2$ + $y^2$ = 2$z^2$
x+y+z=1
We are supposed to find dx/dz, dy/dz, d$^2$x/d$z^2$,d$^2$y/d$z^2$ at x=1, y=-1, z=1.
I am getting confused because in a way I could just write y=1-z-x, making y a function of z and x. A partial differentiation of y and z then wouldn't yield 0 but some form of 2y(dy/dx) and so on. Is this approach wrong? How should I solve it?
Your question has to do with the confusion between total derivatives and partial derivatives. It is not clear from the statement of your question which is to be found. However, as you have pointed out, if you assume it is asking for partial derivatives then the first equation is extraneous, as $\partial x/\partial z$ and $\partial y/\partial z$ are both -1 and the second partials vanish.
I think it is more reasonable to assume that the question is asking for total derivatives. Conceptually, the region of intersection of the cone and the plane is an ellipse in three dimensions, and one could parametrize the intersection using an auxiliary variable $t$ as $x=x(t),y=y(t),z=z(t)$, in which case you could compute the total derivatives as $dx/dz=x'(t)/z'(t)$ and so forth.
However, a simpler computational approach (without having to parametrize the intersection) would be to set up a system of two equations for the unknown functions $dx/dz$ and $dy/dz$. Taking total derivatives with respect to $z$ of the first and second equations respectively gives $$ x\frac{dx}{dz}+y\frac{dy}{dz}=2z,\qquad \frac{dx}{dz}+\frac{dy}{dz}+1=0. $$ Now plug in the values for $x,y,z$ and solve the $2\times 2$ system to conclude. (And similarly for the second derivatives.)