Convert to $f(x,y) = 2$

Similar approach as Ossama Ismail 's.

Consider

$\begin{aligned} (x-1)^2 + (y-1)^2 & = x^2 - 2x + 1 + y^2 - 2y + 1 \\ & = x^2 + y^2 - 2({\color{#3D99F6}x+y}) + 2 & \small \color{#3D99F6} \text{Given that } x+y = x^2 - xy + y^2 \\ & = x^2 + y^2 - 2({\color{#3D99F6}x^2 - xy + y^2}) + 2 \\ & = - x^2 + 2xy - y^2 + 2 \\ & = - (x-y)^2 + 2 \end{aligned}$

$\begin{aligned} \implies (x-1)^2 + (y-1)^2 + (x-y)^2 & = 2 & \small \color{#3D99F6} \text{Let }a = x-1, \ b = y - 1 \\ \implies a^2 + b^2 + (a-b)^2 & = 2 & \small \color{#3D99F6} \implies a - b = x -y \\ a^2 -ab + b^2 & = 1 \\ {\color{#3D99F6}b^2} - a{\color{#3D99F6}b} + a^2-1 & = 0 & \small \color{#3D99F6} \text{A quadratic equation of }b \end{aligned}$

$\begin{aligned} \implies b & = \frac {a \pm \sqrt{a^2-4a^2 + 4}}2 \\ & = \frac {a \pm \sqrt{4-3a^2}}2 \end{aligned}$

For real $b$ , $4-3a^2 \ge 0$ $\implies a = -1, 0, 1$ . Then we have:

$\begin{cases} a = -1 & \implies b = \dfrac {-1 \pm \sqrt{4-3}}2 = 0, -1 & \implies (x,y) = \begin{cases} (0, 1) \\ (0, 0) \end{cases} \\ a = 0 & \implies b = \dfrac {0 \pm \sqrt{4-0}}2 = 1, -1 & \implies (x,y) = \begin{cases} (1, 2) \\ (1, 0) \end{cases} \\ a = 1 & \implies b = \dfrac {1 \pm \sqrt{4-3}}2 = 1, 0 & \implies (x,y) = \begin{cases} (2, 2) \\ (2, 1) \end{cases} \end{cases}$

$\implies \displaystyle \sum_{i=1}^6 (x_i + y_i) = 1+0+3+1+4+3 = \boxed{12}$

The given equation can be converted to $(x-1)^2 + (y-1)^2 +(x-y)^2 =2$

$(x^2 + y^2 - xy) = (x+y)$

$(2x^2 + 2y^2 - 2xy) = 2(x+y)$

$(x^2 -2x +1) + (y^2 - 2y +1) +(x^2 +y^2 -2xy) - 2 = 0$

$(x-1)^2 + (y-1)^2 +(x-y)^2 =2$

It has the solutions: $(0.0), (0,1), (1,0), (1,2), (2,1), \ and \ (2,2)$

My approach is a bit different.

We get

$\large(x+y)= (x+y)^{2} -3xy$

This can be written as $\large (x+y)(x+y-1)=3xy$

Then we can put certain cases like $\large x+y=3, x+y-1=xy$ and we would get the solutions as

$\large (0,0), (1,0),(0,1),(1,2),(2,1),(2,2)$ .

Another thing to observe here is that

$\large a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ .

So essentially the question asks for how many positive integers $\large x,y$

$\large x^{3}+y^{3}=m^{2},$ where "m" is an integer.

The above fact can be used to establish a solution consisting of both $\large x,y$ as negative integers does not exist and at the same time also establish that a solution where one integer can be +ve and other -ve can also not exist.