How to derive rate of product formation for an allosterically inhibited enzyme?
If you apply quasi-steady state approximation for the complexes then you will get three equations for each of the complexes. Then you have to use the conservation law for enzyme to obtain the equation with respect to just substrate and inhibitor. If you try solving these algebraic equations, it gets very complex and difficult to solve by hand. However, in this system of chemical reactions, you can notice one thing: the binding of the inhibitor is not affected by the binding of substrate. In other words, the amount of enzyme inhibited will not depend on the amount of substrate present. So, in the absence of substrate:
$$\frac{d}{dt}(EI)=k_3I.E - k_{-3}EI = 0$$

So the total uninhibited enzyme ($E_u$) would be:
$$E_u=\frac{E_0}{1+\frac{I}{K_I}}$$

When the substrate is present, the amount of total enzyme available for catalysis would be $E_u$. Now, apply the Michaelis-Menten equation for uninhibited enzyme but substitute total enzyme ($E_0$) with total available enzyme ($E_u$):

$$V=\frac{k_2S\left(\frac{E_0}{1+\frac{I}{K_I}}\right)}{K_M+S}$$

From here you can get the equation given in the book.

The above logic works only if the rate of inhibitor binding is much faster than that of substrate binding. If not, the dynamics of product formation will depend on the different reaction rates and this simplistic equation would not be valid.