I was given the following problem:
Let $f(x, y) = \frac{x-y}{x+y}$ and $P=(1, 1, f(1, 1))$. Find the gradient of $f$ at $P$, the equation of the plane tangent to $f$ at $P$, and the equation of the line tangent to the level curve that passes through $P$.
I am very new to multivariate calculus. I was able to find the gradient and the equation for the tangent plane. However, I am confused as to what would the equation of the line tangent to the level curve be.
I was able to determine that the level curve of the surface drawn by $f$ at the given point is $x=y$, since
$$f(1, 1)=0 \space \text{ and } \space \frac{x-y}{x+y}=0 \iff x=y$$
The geometric interpretation of this, as I am beginning to grasp, is that the two-dimensional line $x=y$ depicts all $(x,y)$ points where $f=0$, or rather where the surface $z=f(x, y)$ has "no height". Furthermore, my mental representation is that $x=y$ in 3D is a plane that crosses the surface of $f$ at every point where $f = 0$, while varying indefinitely in $z$. Almost as if the $2D$ line $x=y$ is the aforementioned plane viewed "from above".
I provided all these mental or geometric representations to give a notion of where my understanding of the concept of level curve is, since I just learnt it. Now, to the problem, I fail to recognize what the line tangent to the level curve would be, since the level curve is itself a line.
I reason that, should the level curve $g(x)$ of $f$ be a non-linear function, I would be able to provide the line tangent to it at $P$ in the form of $t = g'(1)x + b$ for some appropriate $b$. Does this mean the line tangent to the level curve $x=y$ is $x=y$ itself? If not, how would one go about this problem?