Substitute $u=\frac{tx}{36}$ (treating $x>0$): $du=\frac{x}{36}dt$, and as $t$ goes from $0$ to $36$, $u$ goes from $0$ to $x$.

${\int }_0^{x}f(u)\cdot \frac{36}{x}du=4\alpha f(x)\Rightarrow {\int }_0^{x}f(u)du=\frac{\alpha x}{9}f(x)$.

Differentiate both sides w.r.t. $x$:

$f(x)=\frac{\alpha }{9}\left[f(x)+x{f}^{\prime }(x)\right]\Rightarrow (9-\alpha )f(x)=\alpha x{f}^{\prime }(x)$.

$\frac{f^{\prime }(x)}{f(x)}=\frac{9-\alpha }{\alpha x}\Rightarrow \ln f(x)=\left(\frac{9}{\alpha }-1\right)\ln x+C$.

$\Rightarrow f(x)=c{x}^{9/\alpha -1}$.

For $f$ to be a standard parabola $y=c{x}^2$, the exponent must equal $2$: $\frac{9}{\alpha }-1=2\Rightarrow \alpha =3$.

$(2,1)$ on the curve: $1=4c\Rightarrow c=\frac{1}{4}$, so $y=\frac{x^2}{4}$.

At $x=-4$: $\beta =\frac{16}{4}=4$.

${\beta }^{\alpha }=4^3=64$.