We are given the function $f: \mathbb{R}^2 \to \mathbb{R}$ defined as follows:

$f(x, y) = x^2 + 2y^2 - 3xy$

The task is to compute the sum of the partial derivatives at the point $(0, 0)$, specifically:

$f_x(0,0) + f_y(0,0) + f_{xy}(0,0)$

Step 1: Compute $f_x(x, y)$ (Partial derivative with respect to $x$)

The partial derivative of $f(x, y)$ with respect to $x$ is: $f_x(x, y) = \frac{\partial}{\partial x}(x^2 + 2y^2 - 3xy) = 2x - 3y$ At $(0, 0)$: $f_x(0, 0) = 2(0) - 3(0) = 0$

Step 2: Compute $f_y(x, y)$ (Partial derivative with respect to $y$)

The partial derivative of $f(x, y)$ with respect to $y$ is: $f_y(x, y) = \frac{\partial}{\partial y}(x^2 + 2y^2 - 3xy) = 4y - 3x$ At $(0, 0)$: $f_y(0, 0) = 4(0) - 3(0) = 0$

Step 3: Compute $f_{xy}(x, y)$ (Mixed partial derivative with respect to $x$ and $y$)

First, we compute $f_{xy}(x, y)$ by differentiating $f_x(x, y) = 2x - 3y$ with respect to $y$: $f_{xy}(x, y) = \frac{\partial}{\partial y}(2x - 3y) = -3$ At $(0, 0)$: $f_{xy}(0, 0) = -3$

Step 4: Sum the derivatives

Now we sum the derivatives: $f_x(0, 0) + f_y(0, 0) + f_{xy}(0, 0) = 0 + 0 - 3 = -3$

Thus, the value of $f_x(0,0) + f_y(0,0) + f_{xy}(0,0)$ is $\boxed{-3}$.

Would you like further details on any of the steps? Here are some related questions for further exploration:

1. How do you compute the second-order mixed partial derivatives for a function of two variables?
2. What conditions must a function satisfy for $f_{xy} = f_{yx}$?
3. How does the chain rule extend to functions of several variables?
4. Can a function of two variables have discontinuous partial derivatives?
5. How does the Hessian matrix help in understanding second-order partial derivatives?

Tip: When computing partial derivatives, always check if mixed derivatives commute, i.e., $f_{xy} = f_{yx}$, using Clairaut's theorem.