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 $-41,-21$