Given that $$S_{r} = \{(x,y,x):x^{2}+y^{2}-z^{2}=r^{2}\}$$ I have constructed a diffeomorphism $f:S_{a} \rightarrow S_{b}$;$$f(x,y,z)=\left(\frac{b}{a}x,\frac{b}{a}y,\frac{b}{a}z\right)$$
and found the tangent spaces ($\textbf{p}\in S_{a}$) $$T_{\textbf{p}} S_{a} =\{X\in\mathbb{R^{3}}:(x,y,-z)\cdot X = 0\}$$ $$T_{f(\textbf{p})} S_{b} =\{Y\in\mathbb{R^{3}}:(x,y,-z)\cdot Y = 0\}.$$
I am now trying to evaluate $df(\bf{p})$ such that we can show $$df(\textbf{p}):T_{\textbf{p}} S_{a}\rightarrow T_{f(\textbf{p})} S_{b}$$
I believe I have to use the fact that $df(\textbf{p})=\frac{d}{dt}(f \circ c)(0)$ for $c(t)$ a smooth curve on $S_{a}$ with $c(0)=p$ and $\frac{d}{dt}c(0)=X$ however I am struggling to compute the derivative such that we get this relation.
As you suggested, choose a curve $c(t) = \langle x(t), y(t), z(t) \rangle$ with $c(0) = \langle x(0), y(0), z(0) \rangle = \mathbf{p}$ and $c'(0) = \langle x'(0), y'(0), z'(0) \rangle = X$. Then, we can differentiate the composition $$ (f \circ c)(t) = \left\langle \frac{b}{a} x(t), \frac{b}{a} y(t), \frac{b}{a} z(t) \right \rangle $$ component-wise to see that
$$ (f \circ c)'(0) = \left\langle \frac{b}{a} x'(0), \frac{b}{a} y'(0), \frac{b}{a} z'(0) \right \rangle = \left\langle \frac{b}{a} X_x, \frac{b}{a} X_y, \frac{b}{a} X_z \right \rangle. $$
Can you check that this resulting vector satisfies the equation of $T_{f(\mathbf{p} )} S_b$?