The given differential equation is:

$$xy \frac{dy}{dx} = (x + 2) (y + 2)$$

To solve this, we first separate the variables. Dividing both sides by $x(y+2)$ (where $x \neq 0$ and $y \neq -2$), we get:

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

Simplifying the integrands, we write the equation as:

$$\int \left( \frac{y+2-2}{y+2} \right) dy = \int \left( 1 + \frac{2}{x} \right) dx$$ $$\int \left( 1 - \frac{2}{y+2} \right) dy = \int 1 \, dx + 2 \int \frac{1}{x} \, dx$$

Integrating both sides, we obtain the general solution:

$$y - 2 \ln|y+2| = x + 2 \ln|x| + C$$

We need to find the particular solution for the curve passing through the point $(1, -1)$. Substituting $x = 1$ and $y = -1$ into the general solution:

$$(-1) - 2 \ln|-1+2| = 1 + 2 \ln|1| + C$$ $$-1 - 2 \ln(1) = 1 + 0 + C$$

Since $\ln(1) = 0$, we get $-1 = 1 + C$, which implies $C = -2$.

Substituting the value of $C = -2$ back into the general solution, the equation of the required curve is:

$$y - 2 \ln|y+2| = x + 2 \ln|x| - 2$$

Or, rearranging:

$$y - x + 2 = 2 \ln|x(y+2)|$$

∴ The solution curve is $\mathbf{y - x + 2 = 2 \ln|x(y+2)|}$.