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How am I supposed to find the equation of a tangent plane on a surface that its equation is not explicit defined in terms of z?

The equation of the surface is: $$ x^{2} -y^{2} -z^{2} = 1 $$

And I need to find all the points where the tangent plane is parallel to the plane: $$z = x + y $$

How can I do that?

My approach at the moment was to think about the gradient vector of the function as being the normal vector of the plane... But the gradient vector of that surface has only two components (partial x and y), and the vector the normal vector of the plane has three components (1,1,-1)...

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My approach at the moment was to think about the gradient vector of the function as being the normal vector of the plane... But the gradient vector of that surface has only two components (partial x and y), and the vector the normal vector of the plane has three components (1,1,-1)...

Don't consider the surface as the graph of one (or more) function(s) of the form $z=g(x,y)$, but as a level set of a function of three variables: $$f(x,y,z) = c$$ where in your case $f(x,y,z)=x^2-y^2-z^2$ and $c=1$. Then the gradient of $f$ is: $$\nabla f = \left( \tfrac{\partial f}{\partial x}, \tfrac{\partial f}{\partial y}, \tfrac{\partial f}{\partial z} \right) = (2x,-2y,-2z)$$and this gradient is perpendicular to the surface, so parallel to a tangent plane at that point.

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    Now it makes sense! So I just need to check when my gradient is going to be a multiple of the normal vector of the plane, right? – 2017-01-05
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    Right, because if those are multiples, they're parallel. In a point _on the surface_, of course. – 2017-01-05
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    *In a point on the surface* : i.e. you have to find the points $(x,y,z)$ such that $\nabla f$ is parallel to $(1,1,-1)$ AND $f(x,y,z)=c$ – 2017-01-05
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    Perpendicular to the surface, so parallel to the tangent plane? Wouldn't it be Parallel to the surface at that point, so parallel to the tangent plane? perpendicular gradient makes no sense to me – 2017-10-22
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One approach would be to calculate the normal to the surface and check when it is parallel to the normal $(1,1,-1)$ of the plane.

The normal of the surface is just the gradient of the implicit function which defines it, i.e. $(2x, -2y, -2z)$.

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    Wait... I agree that the normal of the surface is the gradient of my implicit function. But how am I able to get $-2z$? 2x is partial x in respect to z, and the same for $-2y$. But $-2z$ is the partial of z in respect to?! I'm not getting it. – 2017-01-05