Find a pair of linear cartesian equations for the line which is tangent to both the surfaces (a) $x^2 +y^2 + 2z^2 = 4 $ and (b) $z=e^{(x-y)}$ at point $(1,1,1)$.
My steps:
(1) find the tangent plane vector at (1,1,1) for each equation
(a) $\langle 2x-2,2y-2,4z-4\rangle$
(b) $\langle\,x-1,y-1,z-1\rangle$
(2) I guess I must now use the cross product of the two vectors above to find the vector that lies on both lines? I am a little confused as to where to go from here, would apreciate any help
You're doing fine. You've found the vectors perpendicular to the two surfaces at any point $(x,y,z)$. Now at $P = (1,1,1)$, you can actually find numerical values for those "normal vectors"; the first is $$ n_1 = \pmatrix{2x, 2y, 4z} = \pmatrix{2, 2, 4} $$ You can compute the second.
The cross product $n_1 \times n_2$, and this will be the direction along a line that lies in both planes (because it's perpendicular to both normal-vectors). Let's suppose that cross product is $v = (a, b, c)$. Then your line passes through $(1,1,1)$ in direction $(a, b, c)$. Can you find the equation for this line?
Hint: I suggest finding a parametric equation for the line rather than an implicit one.