Let $f: \mathbb R^3\rightarrow \mathbb R, g:\mathbb R^2\rightarrow \mathbb R$ be differentiable functions with $f(x,y,z)=g(xy,yz)$ and suppose that $g(u,v) $ satisfies $g(2,6)=2, g_u(2,6)=-1, g_v(2,6)=3$.
Show that $f^{-1}(2)$ admits a tangent plane at $(1,2,3)$ and find an equation for it.
First off, what does it mean that a set "admits" a tangent plane? What definition of a tangent plane should I use? (This problem isn't part of any course, so there isn't any "default definition". I know that for manifolds in $\mathbb R^n$ the tangent plane at a point $x$ is the image of the differential at $x$ of any local parametrization at $x$.) Any help?
You are being asked to show that there is a well-defined tangent plane to the level surface $f(x,y,z)=2$ at the point $(1,2,3)$ and to find its equation. Implicit in this question is that $f(1,2,3)=2$. The former is pretty much a given since you’re told that both $f$ and $g$ are differentiable. To find the equation of the tangent plane, use the fact that the gradient of a function is normal to its level surfaces. You’ll need to use the chain rule to compute this gradient.