To solve this system of equations, we can first rearrange the equations to isolate y in terms of x in both of them.

Equation 1: xy - 2y^2 = 43
y(x - 2y) = 43
y = 43 / (x - 2y)

Equation 2: xy - 3y^2 = -37
y(x - 3y) = -37
y = -37 / (x - 3y)

Now we can set the two expressions for y equal to each other:

43 / (x - 2y) = -37 / (x - 3y)

Cross-multiplying, we get:

43(x - 3y) = -37(x - 2y)
43x - 129y = -37x + 74y
80x = 203y
y = 80x / 203

Now we can substitute this expression for y back into either of the original equations to solve for x:

x(80x / 203) - 2(80x / 203)^2 = 43
(80x^2 / 203) - 2(6400x^2 / 203^2) = 43
(80x^2 / 203) - (128000x^2 / 203^2) = 43
203(80x^2) - 128000x^2 = 43(203^2)
203(80x^2 - 128000) = 43(41209)
16240x^2 - 25984000 = 1772427
16240x^2 = 27601427
x^2 = 1697
x = ±41

Therefore, the solutions to the system of equations are x = 41 and x = -41. Let's substitute these values back into one of the original equations to find the corresponding y values.

For x = 41:

y(41) - 2y^2 = 43
41y - 2y^2 = 43
2y^2 - 41y + 43 = 0
This equation has two solutions for y: y = 1 and y = 21

Therefore, when x = 41, the solutions are (41, 1) and (41, 21)

For x = -41:

y(-41) - 2y^2 = 43
-41y - 2y^2 = 43
2y^2 + 41y - 43 = 0
This equation also has two solutions for y: y = -1 and y = -21

Therefore, when x = -41, the solutions are (-41, -1) and (-41, -21)

In conclusion, the solutions to the system of equations xy - 2y^2 = 43 and xy - 3y^2 = -37 are (41, 1), (41, 21), (-41, -1), and (-41, -21)