To solve the equation \(3^{2y} + 6 \cdot 3^y = 27\), let’s proceed step-by-step.

1. Rewrite the equation:
\[
3^{2y} + 6 \cdot 3^y = 27
\]

2. Let \(3^y = x\). Then, \(3^{2y} = x^2\). Substitute these into the equation:
\[
x^2 + 6x = 27
\]

3. Rearrange the equation into a standard quadratic form:
\[
x^2 + 6x - 27 = 0
\]

4. Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 1\), \(b = 6\), and \(c = -27\).

Calculate the discriminant:
\[
b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-27) = 36 + 108 = 144
\]

So:
\[
x = \frac{-6 \pm \sqrt{144}}{2 \cdot 1} = \frac{-6 \pm 12}{2}
\]

5. Find the solutions for \(x\):
\[
x = \frac{-6 + 12}{2} = \frac{6}{2} = 3
\]
\[
x = \frac{-6 - 12}{2} = \frac{-18}{2} = -9
\]

6. Substitute back \(x = 3^y\):
- For \(x = 3\):
\[
3^y = 3 \implies y = 1
\]
- For \(x = -9\), this is not possible since \(3^y\) cannot be negative.

7. Therefore, the valid solution is \(y = 1\).