How to have a horizontal line work inside a tabbing environment

I've used LaTeX for over 25 years and I don't think I've ever really seen a good use for tabbing but anyway I think that you want lines to look like this?

\documentclass[a4paper, 12pt]{article}
\usepackage{mathptmx}

\newcommand{\tab}{\hspace*{1em}}
\makeatletter
\def\linefill{%
\leavevmode
\leaders\hrule\hskip\dimexpr\textwidth -\@tempdima\mbox{}}
\begin{document}
\begin{tabbing}

\textbf{Question 1} \\ 
\textbf{a.} \tab \= Let $y = \left(- 3 x^{2} - 3 x\right)^{3}$. Find $\frac{dy}{dx}$. \\ 
\\ 
\> \linefill \\ 
\> \linefill \\ 
\> \linefill \\ 
\> \linefill \\ 
\> \linefill \\ 
\textbf{b.} \tab Let $f(x) = e^{2 x^{2} + 9 x + 5}$. Evaluate $f'(-1)$. \\ 
\\ 
\> \linefill \\ 
\> \linefill \\ 
\> \linefill \\ 
\> \linefill \\ 
\> \linefill \\ 


Let $f(x) = \left(- 3 x^{2} - 3 x\right)^{3} = u^{3}, u = - 3 x^{2} - 3 x$ \\ 
$f'(x) = 3 u^{2} \times u'$ \\ 
$f'(x) = - 81 x^{2} \left(x + 1\right)^{2} \left(2 x + 1\right)$ \\ 
\\
$f'(x) = \left(4 x + 9\right) e^{2 x^{2} + 9 x + 5}$ \\ 
$f'(-1) = \frac{5}{e^{2}}$ \\ 

\end{tabbing}

\noindent X\dotfill X
\end{document}

I like David's answer, but I couldn't let go the opportunity to use \tikzmark. In this, \tikzmark is used to place a mark at the desired x-position where the lines should begin, and then \DrawLines is used to draw the lines; the syntax is

\DrawLines[<length>]{<number>}{<mark>}

where <number> is a positive integer representing the amount of lines to be drawn, and <mark> is a previously stablished mark set with \tikzmark; <length> allows control over the vertical separation between consecutive lines.

The code:

\documentclass[a4paper, 12pt]{article}
\usepackage{mathptmx}
\usepackage{tikzpagenodes}
\usetikzlibrary{calc}

\newcommand{\tab}{\hspace*{1em}}

\newcounter{tmp}

\makeatletter
\def\whileexpr#1\do#2{%
  #1\expandafter\@iden\else\expandafter\@gobble\fi
  {#2\whileexpr#1\do{#2}}%
}
\makeatother

\newcommand\tikzmark[1]{%
  \tikz[remember picture,overlay]\node[inner xsep=0pt] (#1) {};}
\newcommand\DrawLines[3][\baselineskip]{%
\\
\setcounter{tmp}{0}%
\begin{tikzpicture}[remember picture,overlay]
\foreach \i in {1,...,#2}
{\path[draw] let \p1=(#3), \p2=(current page text area.west) in ([yshift=-\i*#1]\x1,0|-0,\y1-\baselineskip) -- ( $ ([yshift=-\i*#1]current page text area.east|-#3) + (0,-\baselineskip) $ );}
\end{tikzpicture}%
\whileexpr\ifnum\thetmp<#2\relax\do{%
    \>\mbox{}\\[\dimexpr#1-\baselineskip\relax]\stepcounter{tmp}\relax
  }%
\>\mbox{}\\[3pt]
}

\begin{document}

\begin{tabbing}
\textbf{Question 1} \\
\\ 
\textbf{a.} \tab \=\tikzmark{A}Let $y = \left(- 3 x^{2} - 3 x\right)^{3}$. Find $\frac{dy}{dx}$.
\DrawLines{4}{A}%
\\
\textbf{b.} \tab\tikzmark{B}Let $f(x) = e^{2 x^{2} + 9 x + 5}$. Evaluate $f'(-1)$. \\ 
\\ 
\DrawLines[25pt]{7}{B}%
Let $f(x) = \left(- 3 x^{2} - 3 x\right)^{3} = u^{3}, u = - 3 x^{2} - 3 x$ \\ 
$f'(x) = 3 u^{2} \times u'$ \\ 
$f'(x) = - 81 x^{2} \left(x + 1\right)^{2} \left(2 x + 1\right)$ \\ 
\\
$f'(x) = \left(4 x + 9\right) e^{2 x^{2} + 9 x + 5}$ \\ 
$f'(-1) = \frac{5}{e^{2}}$ \\ 
\end{tabbing}

\end{document}

THe code needs two or three runs to stabilize.

I used the \whileexpr macro from Ahmed's Musa's answer to looping in tabular.