Let a curve y = y(x) pass through the point (3,3) and the area of the region under this curve, above the x-axis
and between the abscissae 3 and x(>3) be ${\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^3$. If this curve also passes through the point$(\mathrm{\alpha },6\sqrt{10})$ in
tte first quadrant, then $\mathrm{\alpha }$ is equal to_______

$\begin{array}{l}{\int }_3^x f\left(x\right)dx={\left(\frac{f\left(x\right)}{x}\right)}^3\\ x^3\cdot {\int }_3^x f\left(x\right)dx=f^3\left(x\right)\text{ Differentiate w.r.t. }\mathrm{x}\\ \begin{array}{l}{x}^{3}f\left(x\right)+3x^2\cdot \frac{f^3\left(x\right)}{x^3}=3f^2\left(x\right)f^{\mathrm{\prime }}\left(x\right)\\ \Rightarrow 3y^2\frac{dy}{dx}=x^{3}y+\frac{3y^3}{x}\\ 3xy\frac{dy}{dx}=x^4+3y^2\\ y^2=t\\ \begin{array}{l}\frac{3}{2}\frac{dt}{dx}=x^3+\frac{3t}{x}\\ \frac{dt}{dx}-\frac{2t}{x}=\frac{2x^3}{3}\\ I.F.=\cdot e^{\int -\frac{2}{x}dx}=\frac{1}{x^2}\\ solution of the differential equation is \\ \begin{array}{l}t\cdot \frac{1}{x^2}=\int \frac{2}{3}xdx\\ \frac{y^2}{x^2}=\frac{x^2}{3}+C\\ y^2=\frac{x^4}{3}+C{x}^2\\ Curve passes through \left(3,3\right)\Rightarrow c=-2\\ y^2=\frac{x^4}{3}-2x^2\\ \text{ Which passes through }(\alpha ,6\sqrt{10})\\ \begin{array}{r}\frac{{\alpha }^4-6{\alpha }^2}{3}=360\\ {\alpha }^4-6{\alpha }^2-1080=0\\ \boxed{\alpha =6}\end{array}\end{array}\end{array}\end{array}\end{array}$