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**Soroban** Hello, cupcakelova87!

I would simplify that log first . . .

4xy + ln(x²) + ln(y) .= .7 . .→ . .4xy + 2·ln(x) + ln(y) .= .7

Differentiate: .4x·y' + 4y + 2/x + y'/y .= .0

Then we have: .4x·y' + y'/y .= .-4y - 2/x

Factor: .(4x + 1/y)·y' .= .2(1/x - 2y)

. . . . . . . . . . . 4xy + 1 . . . . . 2(1 - 2xy)

And we have: . --------- · y' .= .-----------

. . . . . . . . . . . . . .y . . . . . . . . . .x

. . . . . . . . . . . . . .2y(1 - 2xy)

Therefore: .y' . = . --------------

. . . . . . - . - . . . . .x(4xy + 1)