Kelvin the Frog and Alex the Kat are playing a game on an initially empty blackboard. Kelvin begins by writing a digit.  Then, the players alternate inserting a digit anywhere into the number currently on the blackboard, including possibly a leading zero (e.g. $12$ can become $123$, $142$, $512$, $012$, etc.). Alex wins if the blackboard shows a perfect square at any time, and Kelvin's goal is prevent Alex from winning. Does Alex have a winning strategy?
1. **Initial Setup and Strategy**:
   - Kelvin and Alex are playing a game where they alternately insert digits into a number on a blackboard.
   - Alex wins if the number on the blackboard becomes a perfect square at any time.
   - Kelvin's goal is to prevent Alex from winning.

2. **Key Claim**:
   - The key claim is that Alex cannot turn both \(10n + 3\) and \(10n + 7\) into perfect squares with some move.
   - To understand this, we need to analyze the properties of numbers modulo 10 and modulo 100.

3. **Modulo 10 Analysis**:
   - Consider the number \(10n + 3\). For this number to become a perfect square, the last digit must be one of the digits that a perfect square can end in. The possible last digits of perfect squares are \(0, 1, 4, 5, 6, 9\).
   - Similarly, consider the number \(10n + 7\). For this number to become a perfect square, the last digit must also be one of \(0, 1, 4, 5, 6, 9\).

4. **Modulo 100 Analysis**:
   - We need to check if there are any perfect squares of the form \(10n + 3\) and \(10n + 7\) modulo 100.
   - The only square of the form \(7\star\) modulo 100 is 76, and the only square of the form \(3\star\) modulo 100 is 36.
   - Therefore, to win, Alex would need to add a 6 at the end of the number to make it a perfect square.

5. **Contradiction**:
   - Suppose both \(a = 100n + 36\) and \(b = 100n + 76\) are perfect squares.
   - The difference between these two numbers is \(b - a = 40\).
   - We need to check if there are any pairs of perfect squares that differ by 40. The only pairs of perfect squares differing by 40 are \(\{81, 121\}\) and \(\{9, 49\}\).
   - Neither of these pairs fit the form \(100n + 36\) and \(100n + 76\), so it is impossible for both to be perfect squares.

6. **Kelvin's Strategy**:
   - On his turn, Kelvin can add either a 3 or a 7 to the end of the current number.
   - The above claim guarantees that one of these two possible moves will prevent the number from becoming a perfect square.
   - Kelvin can repeat this strategy indefinitely, ensuring that Alex never wins.

Conclusion:
\[
\boxed{\text{Kelvin can win.}}
\]