We have
$$\begin{aligned} \frac{x^3 - xy + 1}{x^2 + x - y} &= \frac{y^3 + xy - 1}{y^2 + x - y} \Rightarrow x - \frac{x^2 - 1}{x^2 + x - y} = y + \frac{y^2 - 1}{y^2 + x - y} \\ \Rightarrow x - y &= \frac{x^2 - 1}{x^2 + x - y} + \frac{y^2 - 1}{y^2 + x - y}. \quad (*) \end{aligned}$$
Now we have three cases:
Case 1.
$$x = y \Rightarrow 2\frac{x^2 - 1}{x} = 0 \Rightarrow x^2 = 1 \text{ or } x = 0 \Rightarrow (x,y) = (0,0), (1,1), (-1,-1)$$

Case 2. $x > y$. Let $k = x - y > 0$ so by (*) we have $k = \frac{x^2 - 1}{x^2 + k} + \frac{y^2 - 1}{y^2 + k}$. If $|x| \le 1$ or $|y| \le 1$ by using main equation we get $(x,y) = (-1,0), (1,2), (0,1), (-2,-1)$. Now we can assume $|x|,|y| > 1$
$$x^2 - 1 < x^2 + k \Rightarrow 0 < \frac{x^2 - 1}{x^2 + k} < 1 \text{ and similarly } 0 < \frac{y^2 - 1}{y^2 + k} < 1, \text{ therefore}$$
$$0 < k = \frac{x^2 - 1}{x^2 + k} + \frac{y^2 - 1}{y^2 + k} < 2 \Rightarrow k = 1 \Rightarrow 1 = \frac{x^2 - 1}{x^2 + 1} + \frac{y^2 - 1}{y^2 + 1}. \quad (**)$$

Case 3. $x < y$. Let $t = y - x > 0$ so by (*) we have $t = \frac{x^2-1}{t-x^2} + \frac{y^2-1}{t-y^2}$ and by adding $2 = \frac{t-x^2}{t-x^2} + \frac{t-y^2}{t-y^2}$ to both sides we get $t+2 = \frac{t-1}{t-x^2} + \frac{t-1}{t-y^2}$.
We claim that $t - x^2$ and $t - y^2$ are not positive, simultaneously. Assume by
contrary $t - x^2 > 0, t - y^2 > 0$ then
$$t > x^2, t > y^2 \Rightarrow 2(y - x) = 2t > x^2 + y^2 \Rightarrow (x + 1)^2 + (y - 1)^2 < 2.$$
This contradicts because $|x|,|y|>1$. Therefore at least one of fractions $\frac{t-1}{t-x^2}$ and $\frac{t-1}{t-y^2}$ is less than or equal to 0 and the other less than or equal to $t-1$. So
$$t + 2 = \frac{t - 1}{t - x^2} + \frac{t - 1}{t - y^2} \le 0 + (t - 1) < t.$$
This contradiction shows that case 3 does not have a new solution and these are all the solutions: $(0,0), (1,1), (-1,-1), (-1,0), (1,2), (0,1), (-2,-1)$. □