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Quotient rule on both side

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is it correct?
what do I do next

Good so far.

\[\frac{y-xy'}{y^2}=\frac{xy'-y}{x^2}\]
I would clear the fractions first.

Multiply both sides by x^2y^2.

yes by calcelling the \[y ^{2} x^{2}\] on each side respectively \[x^{2}(y-xy^1)+y^2(xy^1-y)\] then?

You mean equal right not plus?

yes

After I multiply I got\[x^2y-x^3y^\prime =xy^2y^\prime-y^3\]

Right. Continue.
Put y' terms together on one side and then non-y' terms on the other.

\[x^2y+y^3=xy^2y^\prime + x^3 y^\prime\]

then im not sure what to do

Factor the y' out on that one side.

\[x^2y+y^3=y'(xy^2+x^3)\]

Divide both sides by what y' is being multiplied by.

You can reduce your answer.

\[x^2y+y^3/xy^2+x^3\]=\[y^\prime\]

\[y'=\frac{x^2y+y^3}{xy^2+x^3} =\frac{y(x^2+y^2)}{x(y^2+x^2)}\]
Anything cancel?

ok show me

hey 2pie
since we have x/y = y/x
can we say x^2 = y^2 ?

i just did a cross-multiplication

yes

I think the slope can be simplified even more by the way when you guys are done. :)

and then we can differentiate
x^2=y^2
to get 2x =2y dy/dx
can you isolaye dy/dx from this ?

yes by dividing 2y on both sides

and what would u get finally ? as dy/dx ?

dy/dx = x/y

And recall from your initial equation y/x=x/y

wow it was easier than I thought thank you very much for your help and time @myininaya @hartnn :)

thank you Im really glad I joined

Who couldn't pass up free math help?

we too are glad that you joined :)

thanks ;)