A square is inscribed in a circle $x^2 + y^2 + 2x + 2y + 1 = 0$ and its sides are parallel to coordinate axes. Which one of the following is a vertex of the square?

The given equation of the circle is $x^2 + y^2 + 2x + 2y + 1 = 0$.

Rewriting it in standard form:

$(x+1)^2 + (y+1)^2 = 1$

The center of the circle is $C(-1, -1)$ and its radius is $r = 1$.

Since the square is inscribed in the circle and its sides are parallel to the coordinate axes, the diagonals of the square make an angle of $45^{\circ}$ with the coordinate axes.

The distance from the center to any vertex is equal to the radius $r = 1$.

The coordinates of the vertices can be found using the parametric form of a circle:

$x = -1 \pm r \cos(45^{\circ}) = -1 \pm \dfrac{1}{\sqrt{2}}$

$y = -1 \pm r \sin(45^{\circ}) = -1 \pm \dfrac{1}{\sqrt{2}}$

Thus, the four vertices of the square are $\left(-1 \pm \dfrac{1}{\sqrt{2}}, -1 \pm \dfrac{1}{\sqrt{2}}\right)$.

Among the given options, $\left(-1+\dfrac{1}{\sqrt{2}}, -1-\dfrac{1}{\sqrt{2}}\right)$ is one of the vertices.

Answer: $\left(-1+\dfrac{1}{\sqrt{2}}, -1-\dfrac{1}{\sqrt{2}}\right)$