Determine which of the given polynomial equations are valid identities.

To determine whether the given polynomial equations are valid identities, we need to analyze each equation step by step and check if they hold true for all values of the variables involved.

---

Equation 1:

$$
z = x^2 - 4xy + 4y^2
$$

#### Step 1: Simplify the right-hand side.
The expression on the right-hand side is:
$$
x^2 - 4xy + 4y^2
$$
This can be rewritten as a perfect square trinomial:
$$
x^2 - 4xy + 4y^2 = (x - 2y)^2
$$

#### Step 2: Compare with the left-hand side.
The left-hand side is simply \( z \). For this equation to be an identity, it must hold for all values of \( x \) and \( y \). Rewriting the equation, we have:
$$
z = (x - 2y)^2
$$
This equation is not an identity because \( z \) is not defined in terms of \( x \) and \( y \) in a general sense. Instead, it expresses \( z \) as a function of \( x \) and \( y \). Therefore, this is not an identity but rather a relationship between \( z \), \( x \), and \( y \).

#### Conclusion for Equation 1:
This is not a valid identity.

---

Equation 2:

$$
-m^2 = 1
$$

#### Step 1: Analyze the equation.
The equation states that the negative of \( m^2 \) equals 1. Recall that \( m^2 \) is always non-negative for any real number \( m \) (i.e., \( m^2 \geq 0 \)). Therefore, \( -m^2 \) is always non-positive (i.e., \( -m^2 \leq 0 \)).

#### Step 2: Check if the equation can hold.
For the equation \( -m^2 = 1 \) to hold, \( -m^2 \) would need to be positive, which is impossible since \( -m^2 \leq 0 \) for all real \( m \).

#### Conclusion for Equation 2:
This equation is never true for any real value of \( m \). Therefore, it is not a valid identity.

---

Equation 3:

$$
k(k + 1) = 3k^2 + 3k
$$

#### Step 1: Expand the left-hand side.
The left-hand side is:
$$
k(k + 1) = k^2 + k
$$

#### Step 2: Compare with the right-hand side.
The right-hand side is:
$$
3k^2 + 3k
$$

#### Step 3: Set the two sides equal and simplify.
Equating the two sides:
$$
k^2 + k = 3k^2 + 3k
$$
Rearrange all terms to one side:
$$
k^2 + k - 3k^2 - 3k = 0
$$
Simplify:
$$
-2k^2 - 2k = 0
$$
Factor out the common term:
$$
-2k(k + 1) = 0
$$

#### Step 4: Solve for \( k \).
The equation \( -2k(k + 1) = 0 \) implies:
$$
k = 0 \quad \text{or} \quad k + 1 = 0 \implies k = -1
$$
Thus, the equation holds only for \( k = 0 \) or \( k = -1 \).

#### Conclusion for Equation 3:
Since the equation does not hold for all values of \( k \) (it is true only for specific values \( k = 0 \) and \( k = -1 \)), it is not a valid identity.

---

Final Answer:

None of the given equations are valid identities. The final answer is:
$$
\boxed{\text{None}}
$$