[latex]\begin{array}{llll}\frac{d}{dx}(x^3 \sin y+y) = \frac{d}{dx}(4x+3) & & & \text{Step 1: Differentiate both sides of the equation.} \\ \frac{d}{dx}(x^3 \sin y)+\frac{d}{dx}(y) = 4 & & & \begin{array}{l}\text{Step 1.1: Apply the sum rule on the left.} \\ \text{On the right,} \, \frac{d}{dx}(4x+3)=4. \end{array} \\ (\frac{d}{dx}(x^3) \cdot \sin y+\frac{d}{dx}(\sin y) \cdot x^3) + \frac{dy}{dx} = 4 & & & \begin{array}{l}\text{Step 1.2: Use the product rule to find} \\ \frac{d}{dx}(x^3 \sin y). \, \text{Observe that} \, \frac{d}{dx}(y)=\frac{dy}{dx}. \end{array} \\ 3x^2 \sin y+(\cos y\frac{dy}{dx}) \cdot x^3 + \frac{dy}{dx} = 4 & & & \begin{array}{l}\text{Step 1.3: We know} \, \frac{d}{dx}(x^3)=3x^2. \, \text{Use the} \\ \text{chain rule to obtain} \, \frac{d}{dx}(\sin y)= \cos y\frac{dy}{dx}. \end{array} \\ x^3 \cos y\frac{dy}{dx}+\frac{dy}{dx} = 4-3x^2 \sin y & & & \begin{array}{l}\text{Step 2: Keep all terms containing} \, \frac{dy}{dx} \, \text{on the} \\ \text{left. Move all other terms to the right.} \end{array} \\ \frac{dy}{dx}(x^3 \cos y+1) = 4-3x^2 \sin y & & & \text{Step 3: Factor out} \, \frac{dy}{dx} \, \text{on the left.} \\ \frac{dy}{dx} = \large \frac{4-3x^2 \sin y}{x^3 \cos y+1} & & & \begin{array}{l}\text{Step 4: Solve for} \, \frac{dy}{dx} \, \text{by dividing both sides of} \\ \text{the equation by} \, x^3 \cos y+1. \end{array} \end{array}[/latex]