Step 1: Form the differential equation. The slope of tangent at any point $(x, y)$ on the curve is $\frac{dy}{dx}$. Given that the product of slope and $y$-coordinate equals the $x$-coordinate:

$$y \cdot \frac{dy}{dx} = x$$

Step 2: Solve by separation of variables. Rewriting:

$$y \, dy = x \, dx$$

Step 3: Integrate both sides.

$$\int y \, dy = \int x \, dx$$

which gives:

$$\frac{y^2}{2} = \frac{x^2}{2} + C$$

Step 4: Find the particular solution. Since the curve passes through $(0, -2)$, substitute $x = 0$ and $y = -2$:

$$\frac{(-2)^2}{2} = \frac{0^2}{2} + C$$

giving $C = 2$.

Step 5: Write the final equation.

$$\frac{y^2}{2} = \frac{x^2}{2} + 2$$

or

$$y^2 = x^2 + 4$$

or

$$y^2 - x^2 = 4$$

∴ The equation of the required curve is $\mathbf{y^2 - x^2 = 4}$.