Transfer function of complicated circuit
Once you find \$\dfrac{v_L}{v_1}\$, then just rearranging the terms will give you the standard form. Let me do this example for you. 

Step 1: Calculate \$\dfrac{v_L}{v_1}\$

From the circuit, the following three equations can be derived directly.
$$v_L = k_3v_3 \frac{R_L}{R_L + R_{s3}} \tag{1}$$
$$v_3 = -g_{m2}v_2(r_{\pi3}||\dfrac{1}{sc_{\pi3}})\tag{2}$$
$$v_2 = -(g_{m2}v_2 - g_{m1}v_1)(r_{\pi2}||\dfrac{1}{sc_{\pi2}})\tag{3.a}$$
$$ \Rightarrow v_2 = \frac{g_{m1}v_1(r_{\pi2}||\dfrac{1}{sc_{\pi2}})}{1+g_{m2}(r_{\pi2}||\dfrac{1}{sc_{\pi2}})}\tag3$$

Step 2: Rearrange the terms

Now from equation (1), (2) and (3), the value of \$v_L\$ can be expressed in terms of \$v_1\$. From this \$\dfrac{v_L}{v_1}\$ can be calculated. The answer will be in the following form:

$$\frac{v_L}{v_1} = K_1 \times \frac{r_{\pi3}}{1+sc_{\pi3}r_{\pi3}}\times  \frac{r_{\pi2}}{1+sc_{\pi2}r_{\pi2}+g_{m2}r_{\pi2}}\tag4$$
where $$K_1= -g_{m1}g_{m2}k_3v_3 \frac{R_L}{R_L + R_{s3}}$$

Now its just a matter of rearranging the terms to convert it into standard form. 
Dividing numerator and denominator of equation (4) by \$(1+g_{m2}r_{\pi2})\$,
$$\Rightarrow \frac{v_L}{v_1} = K_1 \times \frac{r_{\pi3}}{1+sc_{\pi3}r_{\pi3}}\times  \frac{\frac{r_{\pi2}}{(1+g_{m2}r_{\pi2})}}{1+s\frac{c_{\pi2}r_{\pi2}}{(1+g_{m2}r_{\pi2})}}$$

$$\Rightarrow \frac{v_L}{v_1} = K \times \frac{1}{1+sc_{\pi3}r_{\pi3}}\times  \frac{1}{1+s\frac{c_{\pi2}r_{\pi2}}{(1+g_{m2}r_{\pi2})}}\tag5$$
where $$K = K_1 \times r_{\pi3} \times \frac{r_{\pi2}}{(1+g_{m2}r_{\pi2})}$$

Now equation (5) is in standard form and value of poles can be calculated directly.