To find the vertex of the parabola given by the equation \( y = x^2 + \frac{13}{2} \), we can start by recognizing that this is a standard form of a parabola \( y = ax^2 + bx + c \). In this case, \( a = 1 \), \( b = 0 \), and \( c = \frac{13}{2} \).
For any parabola in this form, the x-coordinate of the vertex can be calculated using the formula:
\[ x = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ x = -\frac{0}{2 \times 1} = 0 \]
Now, substitute \( x = 0 \) back into the equation to find the y-coordinate:
\[ y = (0)^2 + \frac{13}{2} = \frac{13}{2} \]
Thus, the vertex of the parabola is:
\[ \left(0, \frac{13}{2}\right) \]
Therefore, the coordinates of the vertex are:
\[ \boxed{\left(0, \frac{13}{2}\right)} \]