Step 1: Find the slope of the line segment joining point $(x, y)$ to point $(-4, -3)$.

$$\text{slope} = \frac{y - (-3)}{x - (-4)} = \frac{y + 3}{x + 4}$$

Step 2: Set up the differential equation. Since slope of tangent is twice the slope of line segment:

$$\frac{dy}{dx} = 2 \times \frac{y + 3}{x + 4}$$

Step 3: Separate variables and integrate. Rearranging:

$$\frac{dy}{y + 3} = \frac{2\,dx}{x + 4}$$

Integrating both sides:

$$\ln|y + 3| = 2\ln|x + 4| + C_1$$

Step 4: Simplify the equation. This gives:

$$y + 3 = C(x + 4)^2$$

where $C = e^{C_1}$.

Step 5: Find constant C using given condition. Since curve passes through $(-2, 1)$:

$$1 + 3 = C(-2 + 4)^2$$ $$4 = C(4)$$ $$C = 1$$

Step 6: Write the final equation. Substituting $C = 1$:

$$y + 3 = (x + 4)^2$$

∴ $\mathbf{y = x^2 + 8x + 13}$