Step 1: Write down the functional equation.
We are given $3f(\cos x) + 2f(\sin x) = 5x$. We want to find $f'(\cos x) + f'(\sin x)$.
Step 2: Differentiate the given equation with respect to $x$.
Differentiating both sides: $3f'(\cos x) \cdot (-\sin x) + 2f'(\sin x) \cdot \cos x = 5$. So we have \[-3\sin x \cdot f'(\cos x) + 2\cos x \cdot f'(\sin x) = 5. \quad \text{...(1)}\]
Step 3: Substitute $x \to \pi/2 - x$ in the original equation.
Replace $x$ with $\pi/2 - x$: $3f(\sin x) + 2f(\cos x) = 5(\pi/2 - x)$. Differentiate with respect to $x$: $3f'(\sin x)\cos x + 2f'(\cos x)(-\sin x) = -5$. So \[3\cos x \cdot f'(\sin x) - 2\sin x \cdot f'(\cos x) = -5. \quad \text{...(2)}\]
Step 4: Solve equations (1) and (2) for $f'(\cos x)$.
Multiply (1) by 3 and (2) by 2: $-9\sin x \cdot f'(\cos x) + 6\cos x \cdot f'(\sin x) = 15$ and $-4\sin x \cdot f'(\cos x) + 6\cos x \cdot f'(\sin x) = -10$. Subtracting the second from the first: $-5\sin x \cdot f'(\cos x) = 25$, giving $f'(\cos x) = -\frac{5}{\sin x}$.
Step 5: Find $f'(\sin x)$.
Substitute back into equation (1): $-3\sin x \cdot (-5/\sin x) + 2\cos x \cdot f'(\sin x) = 5$, so $15 + 2\cos x \cdot f'(\sin x) = 5$, giving $2\cos x \cdot f'(\sin x) = -10$, so $f'(\sin x) = -\frac{5}{\cos x}$.
Step 6: Add the two results.
$f'(\cos x) + f'(\sin x) = -\frac{5}{\sin x} - \frac{5}{\cos x}$. \[\boxed{-\dfrac{5}{\sin x} - \dfrac{5}{\cos x}}\]