= Solution
The <contraction mapping theorem> says that if $(X,d)$ is a nonempty <complete metric space> and $T:X\to X$ satisfies $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$, then $T$ has exactly one <fixed point>.
For the proof, start at $x_0\in X$ and set $x_n=T^nx_0$. Iterating the <contraction> bound gives $d(x_{n+1},x_n)\le q^nd(x_1,x_0)$. The <triangle inequality> then gives, for $m>n$,
$$
d(x_m,x_n)\le\sum_{k=n}^{m-1}q^kd(x_1,x_0)\le\frac{q^n}{1-q}d(x_1,x_0).
$$
Thus $(x_n)$ is a <Cauchy sequence> and has a limit $x_*$ by completeness. Since the <contraction> is continuous, $Tx_*=\lim Tx_n=\lim x_{n+1}=x_*$. If $y_*$ is another <fixed point>, then $d(x_*,y_*)\le qd(x_*,y_*)$, forcing distance zero. This proves existence and uniqueness.
For the <integral> map, the <fundamental theorem of calculus> shows that $Tf$ is continuous whenever $f$ is continuous. In the <supremum norm>,
$$
|Tf(x)-Tg(x)|\le3\|f-g\|_\infty\left|\int_0^x|t|\,dt\right|=\frac32x^2\|f-g\|_\infty.
$$
Hence $\|Tf-Tg\|_\infty\le\frac32\max(a^2,b^2)\|f-g\|_\infty$. We can choose
$$
\boxed{a=-\frac12,\quad b=\frac12,\quad q=\frac38.}
$$
The space $C[a,b]$ with this <supremum norm> is complete: a uniformly <Cauchy sequence> has pointwise real limits, the Cauchy bound passes to those limits to give <uniform convergence>, and a <uniform limit> of <continuous functions> is continuous. The <contraction mapping theorem> therefore gives a unique continuous $y$ with $y(x)=1+\int_0^x3ty(t)\,dt$. The <fundamental theorem of calculus> makes this $y$ differentiable, with $y'=3xy$ and $y(0)=1$. Conversely every solution of that initial-value problem obeys the <integral> equation, so uniqueness transfers to the <ordinary differential equation>. Explicitly \b[$y(x)=e^{3x^2/2}$ on $[-1/2,1/2]$], as direct <differentiation> confirms.
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