As a first step, we use the fact that this is a quadratic in y:
-y²(5x+3x³)+6x²y-9=0 or y²(5x+3x³)-6x²y+9=0.
y²-6xy/(5+3x²)+9/(5x+3x³)=0,
y²-6xy/(5+3x²)+(3x/(5+3x²))²=(3x/(5+3x²))²-9/(5x+3x³),
(y-3x/(5+3x²))²=(9/(5+3x²)(x²/(5+3x²)-1/x),
(y-3x/(5+3x²))²=(3/(5+3x²))²((x³-3x²-5)/x);
y=3x/(5+3x²)±(3/(5+3x²)√((x³-3x²-5)/x).
By observation, if x=-1, the right-hand side becomes a perfect square, because (x³-3x²-5)/x=9.
Substitute x=-1 in the original equation:
8y²+6y-9=0=(4y-3)(2y+3) and y=¾ or -³⁄₂.
Therefore x=-1 and y=¾ or x=-1 and y=-³⁄₂ is a solution which gives us a clue about the factors.
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