Given,

$f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$

Now, solving $\mathrm{L}.\mathrm{H}.\mathrm{L}.$ at $x=\frac{\mathrm{\pi }}{2}$ we get,

$\underset{x\to \frac{{\pi }^{+}}{2}}{\lim }{e}^{\frac{\cot 6x}{\cot 4x}}=\underset{x\to \frac{{\pi }^{+}}{2}}{\lim }{e}^{\frac{\sin 4x·\cos 6x}{\sin 6x·\cos 4x}}=e^{2/3}$

Similarly, on simplification for $\mathrm{R}.\mathrm{H}.\mathrm{L}.$ we get

$\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }{\left(1+\left|\cos x\right|\right)}^{\frac{\lambda }{\left|\cos x\right|}}=e^{\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }\lambda }=e^{\lambda }$

$\because f\left(\frac{\pi }{2}\right)=\mu$

For continuous function, 

$f\left({\frac{\pi }{2}}^{-}\right)=f\left(\frac{\pi }{2}\right)=f\left({\frac{\pi }{2}}^{+}\right)$

$\Rightarrow {\mathrm{e}}^{\frac{2}{3}}={\mathrm{e}}^{\lambda }=\mu$

$\Rightarrow \lambda =\frac{2}{3}, \mu ={\mathrm{e}}^{\frac{2}{3}}$

Now,

$9\lambda +6 {\log }_e\mu +{\mu }^6-{\mathrm{e}}^{6\lambda }$

$=9\left(\frac{2}{3}\right)+6\left(\frac{2}{3}\right)+{\left(e^{\frac{2}{3}}\right)}^6-e^{6\left(\frac{2}{3}\right)}$

$=6+4+e^4-e^4$

$=10$