Given the scalar function f(x,y,z) = x² + y.

The gradient of a scalar function $f$ is given by:

$$\nabla f = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j} + \frac{\partial f}{\partial z} \mathbf{k}$$

First, calculate the partial derivatives:

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 + y) = 2x$$ $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 + y) = 1$$ $$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z}(x^2 + y) = 0$$

Substitute these into the gradient formula:

$$\nabla f = 2x \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k}$$ $$\nabla f = 2x \mathbf{i} + \mathbf{j}$$

The gradient of $f$ is $\boxed{\mathbf{2x \mathbf{i} + \mathbf{j}}}$.

Divergence is defined for a vector field. Assuming the question asks for the divergence of the gradient of $f$, which is $\nabla \cdot (\nabla f)$.
Let $\mathbf{F} = \nabla f = 2x \mathbf{i} + \mathbf{j} + 0 \mathbf{k}$.
The divergence of a vector field $\mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k}$ is given by:

$$\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$$

Here, $P = 2x$, $Q = 1$, and $R = 0$.
Calculate the partial derivatives:

$$\frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(2x) = 2$$ $$\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(1) = 0$$ $$\frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(0) = 0$$

Substitute these into the divergence formula:

$$\nabla \cdot (\nabla f) = 2 + 0 + 0 = 2$$

The divergence of the gradient of $f$ is $\boxed{\mathbf{2}}$.

Curl is defined for a vector field. Assuming the question asks for the curl of the gradient of $f$, which is $\nabla \times (\nabla f)$.
Let $\mathbf{F} = \nabla f = 2x \mathbf{i} + \mathbf{j} + 0 \mathbf{k}$.
The curl of a vector field $\mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k}$ is given by:

$$\nabla \times \mathbf{F} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right) \mathbf{i} + \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right) \mathbf{j} + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \mathbf{k}$$

Here, $P = 2x$, $Q = 1$, and $R = 0$.
Calculate the partial derivatives:

$$\frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(0) = 0$$ $$\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(1) = 0$$ $$\frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(2x) = 0$$ $$\frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(0) = 0$$ $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(1) = 0$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2x) = 0$$

Substitute these into the curl formula:

$$\nabla \times (\nabla f) = (0 - 0) \mathbf{i} + (0 - 0) \mathbf{j} + (0 - 0) \mathbf{k} = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}$$ $$\nabla \times (\nabla f) = \mathbf{0}$$

The curl of the gradient of $f$ is $\boxed{\mathbf{0}}$.