Let $f(x)$ be a differentiable function satisfying $$f(x)=\frac{2}{\sqrt{3}}\int_{0}^{\sqrt{3}}f\left(\frac{\lambda^2x}{3}\right)d\lambda,x>0$$ and $f(1)=\sqrt3$ .
If $y=f(x)$ passes through the point $(\alpha,6)$ then determine the value of $\alpha$ .
My Attempt
Since $f(1)=\sqrt3$ we have $\int_{0}^{\sqrt3}f(\frac{\lambda^2}{3})d\lambda=\frac{3}{2}$ and since $f(\alpha)=6$ we have $\int_{0}^{\sqrt3}f(\frac{\lambda^2\alpha}{3})d\lambda=3\sqrt3$.
Also it is given that $f(x)$ is differentiable so $f'(x)=\frac{2}{3\sqrt3}\int_{0}^\sqrt3f'(\frac{\lambda^2x}{3})\lambda^2d\lambda$.
After this I am not able to proceed.
Note that $$f(x)=\frac{2}{\sqrt{3}}\int_{0}^{\sqrt{3}}f\left(\frac{\lambda^2x}{3}\right)d\lambda\overset{\frac{\lambda^2x}{3}\to \lambda}=\frac{1}{\sqrt{x}}\int_{0}^{x}f(\lambda)\frac{1}{\sqrt{\lambda}}d\lambda$$ and hence $$ (\sqrt xf(x))'=f(x)\frac{1}{\sqrt{x}}=\frac{\sqrt{x}f(x)}{x}. $$ So $$ \frac{(\sqrt xf(x))'}{\sqrt{x}f(x)}=\frac1x$$ or $$\ln(\sqrt{x}f(x))=\ln x+C$$ or $$ \sqrt{x}f(x)=Cx $$ or $f(x)=C\sqrt x$. By $f(1)=\sqrt3$, one has $C=\sqrt3$ and hence $$ f(x)=\sqrt{3x}. $$