Given:
\[
y = \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right) + \tan^{-1} \left( \frac{4x}{1 + 5x^2} \right)
\]
To find \( \frac{dy}{dx} \), we differentiate the equation with respect to \( x \).
Step 1: Derivative of arctangent
The derivative of \( \tan^{-1}(z) \) with respect to \( x \) is:
\[
\frac{d}{dx} \left( \tan^{-1}(z) \right) = \frac{1}{1 + z^2} \cdot \frac{dz}{dx}
\]
Step 2: Differentiate the first term
Let \( z_1 = \frac{2 + 3x}{3 - 2x} \).
Differentiate \( z_1 \) with respect to \( x \):
\[
\frac{dz_1}{dx} = \frac{(3 - 2x)(3) - (2 + 3x)(-2)}{(3 - 2x)^2}
\]
Simplifying:
\[
\frac{dz_1}{dx} = \frac{9 + 6x + 4 + 6x}{(3 - 2x)^2} = \frac{13 + 12x}{(3 - 2x)^2}
\]
The derivative of the first term is:
\[
\frac{1}{1 + z_1^2} \cdot \frac{dz_1}{dx} = \frac{1}{1 + \left( \frac{2 + 3x}{3 - 2x} \right)^2} \cdot \frac{13 + 12x}{(3 - 2x)^2}
\]
Step 3: Differentiate the second term
Let \( z_2 = \frac{4x}{1 + 5x^2} \).
Differentiate \( z_2 \) with respect to \( x \):
\[
\frac{dz_2}{dx} = \frac{(1 + 5x^2)(4) - 4x(10x)}{(1 + 5x^2)^2} = \frac{4 + 20x^2 - 40x^2}{(1 + 5x^2)^2} = \frac{4 - 20x^2}{(1 + 5x^2)^2}
\]
The derivative of the second term is:
\[
\frac{1}{1 + z_2^2} \cdot \frac{dz_2}{dx} = \frac{1}{1 + \left( \frac{4x}{1 + 5x^2} \right)^2} \cdot \frac{4 - 20x^2}{(1 + 5x^2)^2}
\]
Step 4: Combine the results
The expression for \( \frac{dy}{dx} \) simplifies to:
\[
\frac{dy}{dx} = \frac{1}{1 + 25x^2}
\]
Thus, the correct answer is:
\[
\boxed{\frac{1}{1 + 25x^2}}
\]