Per Unit Value Vs Substitution

What if we were to use,

$\rho_n=\cfrac{8\pi}{\lambda^4}$

why is this different from,

$\rho_n=\cfrac{8\pi}{c^3}f^2$

Because in actual,

$\rho_{n\,\lambda}=\cfrac{dn_s}{d\lambda}=\cfrac{8\pi}{\lambda^4}$

and

$\rho_{n\,f}=\cfrac{dn_s}{df}=\cfrac{8\pi}{c^3}f^2$

Because,

$f\lambda=c$

$df\lambda+fd\lambda=0$

$df=-\cfrac{f}{\lambda}d\lambda$

and

$\rho_{n\,\lambda}*\cfrac{d\lambda}{df}=\cfrac{dn_s}{d\lambda}\cfrac{d\lambda}{df}=\cfrac{dn_s}{df}=\rho_{n\,f}$

So,

$\rho_{n\,f}=\cfrac{8\pi}{\lambda^4}*\cfrac{\lambda}{f}=\cfrac{8\pi}{\lambda^3}*\cfrac{1}{f}$

$\rho_{n\,f}=\cfrac{8\pi}{c^3}*\cfrac{f^3}{f}=\cfrac{8\pi}{c^3}f^2$

It is a mistake to just substitute for $\lambda=\cfrac{c}{f}$

$\rho_n=\cfrac{8\pi}{\lambda^4}=\cfrac{8\pi}{\left(\cfrac{c}{f}\right)^4}=\cfrac{8\pi}{c^4}f^4\ne\rho_{n\,f}$

Have a nice day.