So the problem is quite simple, but I want to make sure I'm doing it right.
Determine the tangent plane of the function $h(x,y) = \frac{x^2-y^2}{x^2+y^2}$ in the point $(1,1,0)$.
My solution:
First, we start by computing the gradient of $h$ by means of the quotient rule: $\begin{align}\nabla h &= \frac1{(x^2+y^2)^2}(2x(x^2+y^2)-(x^2-y^2)2x,\; -2y(x^2+y^2)+(x^2-y^2)2y)\\ &=\frac{4y^2}{(x^2+y^2)^2}(x,\; -y)\end{align}$
Then, by filling in to this gradient the coordinates $(1,1)$, we obtain $\nabla h(1,1) = (1,-1)$.
From here we take two different vectors $(0,1), (1,0)$ so that when we move in the $y$ direction by a step of $1$ over the tangent plane, we change $z$ by $(0,1)\cdot(1,-1) = -1$ And analogous for $x$, we change by $1$.
Hence, the tangent plane is the plane described by $z = x-y$.
In particular (aside from the correctness) I'm not sure how to move from gradient and directional change to the tangent plane. The way I'm describing it here doesn't seem very... clean..
The easierst way to determine the tangent plane to a function is via the partial derivatives. You can compute them directly, or via the gradient. If your gradient at a specific point $(x_0,y_0,z_0)$ is the vector $\nabla h(x_0,y_0)=(a,b)$, then the partial dertivatives are
$$\partial_xh(x_0,y_0)=\nabla h(x_0,y_0)\cdot \underbrace{(1,0)}_{x\text{-direction}}=(a,b)\cdot(1,0)=a.$$
In the same way you find $\partial_yh(x_0,y_0)=b$.
These partial derivatives give you two spanning vectors of the tangent plane, namely $(a,0,1)$ and $(0,b,1)$. Via these, you can compute the normal vector $n=(a,0,1)\times (0,b,1)=(-b,-a,ab)$. The tangent plane then is
$$-bx-ay+abz=d.$$
We have to find $d$ from the information that the plane goes through $(x_0,y_0,z_0)$:
$$-bx_0-ay_0+abz_0=d\quad\implies\quad\text{plane equation: }\; -bx-ay+abz=-bx_0-ay_0+abz_0.$$
We can write this in a nicer form:
$$z=\frac{b(x-x_0)+a(y-y_0)+abz_0}{ab}=z_0+\frac1a(x-x_0)+\frac1b(y-y_0).$$
In your case, with $(x_0,y_0,z_0)=(1,1,0)$ and $(a,b)=(1,-1)$, this gives
$$z=(x-1)-(y-1)=x-y.$$