We are given:
\[
a(4 + x^2) = x + y - x^3
\]
\[
x + y - x^3 = a^3 \frac{dy}{dx}
\]
Step 1: Differentiate both sides of the first equation implicitly w.r.t. \(x\):
\[
\frac{d}{dx}[a(4 + x^2)] = \frac{d}{dx}[x + y - x^3]
\Rightarrow a(2x) = 1 + \frac{dy}{dx} - 3x^2
\]
Step 2: Rearranging the above:
\[
\frac{dy}{dx} = a(2x) - 1 + 3x^2 \quad \cdots (1)
\]
Step 3: From original equation:
\[
x + y - x^3 = a^3 \frac{dy}{dx}
\quad \cdots (2)
\]
Put \( x = 1 \) into equation (1):
\[
\frac{dy}{dx} = a(2 \cdot 1) - 1 + 3(1)^2 = 2a + 2 \quad \cdots (3)
\]
From original equation, at \( x = 1 \):
\[
a(4 + 1^2) = 1 + y - 1 \Rightarrow 5a = y \quad \cdots (4)
\]
Now put in equation (2):
\[
x + y - x^3 = a^3 \frac{dy}{dx} \Rightarrow 1 + 5a - 1 = a^3 \cdot (2a + 2)
\Rightarrow 5a = a^3(2a + 2)
\]
Divide both sides by \( a \neq 0 \):
\[
5 = a^2(2a + 2) = 2a^3 + 2a^2
\Rightarrow 2a^3 + 2a^2 - 5 = 0
\]
Solving this cubic equation, we try rational root \( a = 1 \):
\[
2(1)^3 + 2(1)^2 - 5 = 2 + 2 - 5 = -1 \neq 0
\]
\( a = 1 \) : \( a = 1 \Rightarrow \frac{dy}{dx} = 2a + 2 = 4 \)