How to approximate $y=\frac{W(e^{cx+d})}{W(e^{ax+b})}$?  How to approximate
$$y=\frac{W(e^{cx+d})}{W(e^{ax+b})}$$
with (a) simple function(s)? 
given $a=-1/\lambda_0$, $b=(\mu_0+\lambda_0)/\lambda_0$, $c=1/\lambda_1$, $d=(\mu_1+\lambda_1-1)/\lambda_1$ for positive $\mu_0,\lambda_0,\mu_1,\lambda_1$
where $W$ is a Lambert $W$ function, i.e., if $y=xe^x$ then $x=W(y)$ 
My problem is that I can not invert the function and get $x=f(y)$ alone and decided to go for some nice approximations.
Thanks alot for any help.

You could try to estimate $W(x)$ by using Newton-Raphson iteration, because $W(c)$ is the root of $x\exp(x)-c$:
$$x_{n+1}= x_n-\frac{f(x_n)}{f'(x_n)} = x_n-\frac{x_n \exp (x_n)-c}{\exp(x_n)(x_n+1)} = \frac{c\exp(-x_n)+x_n^2}{x_n+1}$$
and as $n$ is sufficiently large, we can get approximations for $W$ using only elementary functions.
Using this method, we can find $W(1)=\Omega = 0.567143\cdots$, starting with $x_0=1$ and keeping 6 places precision:
Iteration       Value         Error
1               0.683939      0.116795
2               0.577454      0.010310
3               0.56723       0.000086
4               0.567143      0.000000

