Answer
$(a) \; y^{'}=\Large\frac{9x}{y}$
$(b) \; y^{'}=\Large\frac{9x}{\sqrt{9x^2-1}}$
$(c) \; $See below
Work Step by Step
$(a)\; 9x^2-y^2=1$ ___(1)
Differentiating (1) implicitly with respect to $x$
$18x-2yy^{'}=0$
$y^{'}=\Large\frac{18x}{2y}=\frac{9x}{y}$ ___(2)
$(b)$ Solving (1) for $y$ in terms of $x$
$y^2=9x^2-1$
$y=\sqrt{9x^2-1}$ ___(3)
Differentiating (3) with respect to $x$
$y^{'}=\frac{18x}{2\sqrt{9x^2-1}}=\frac{9x}{\sqrt{9x^2-1}}$
$(c)\; $ Substitute $y=\sqrt{9x^2-1}$ from (3) in (2)
$y^{'}=\frac{9x}{\sqrt{9x^2-1}}$
Therefore our solutions to parts $(a)$ and $(b)$ are consitent.