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I have moved your topic to the Calculus sub-forum, as it involves differentiation.

Can you show what you have tried so far? You want to implicitly differentiate with respect to $x$. This means you are going to need to apply the product, power and chain rules. Can you find \(\displaystyle \frac{dy}{dx}\)?

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You now want to simplify by canceling or dividing out factors common to the numerator and denominator.

I appreciate the clarification, because initially I did not know where $a$ and $b$ came from, until I found \(\displaystyle \frac{dy}{dx}\).

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We are given:

\(\displaystyle x^3y^2=128\)

Implicitly differentiating with respect to $x$, we find:

\(\displaystyle x^32y\frac{dy}{dx}+3x^2y^2=0\)

\(\displaystyle \frac{dy}{dx}=-\frac{3x^2y^2}{2x^3y}=-\frac{3y}{2x}\)

Now, we are told to equate this to 3 at the point $(a,b)$:

\(\displaystyle \left.\frac{dy}{dx}\right|_{(x,y)=(a,b)}=3\)

\(\displaystyle -\frac{3b}{2a}=3\)

Can you proceed?

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\(\displaystyle \frac{dy}{dx}=-\frac{3y}{2x}\)

so substituting this result back into it will only result in an identity as I showed above.

You now have that \(\displaystyle a=2\), so what is $b$?

By the way, you will get much better help if you give the problems exactly as stated in their entirety up front. Giving the entire problem, along with the directions, and all work you have done is the best way to get prompt, pertinent help.

I'm not "fussing," but rather offering you helpful advice.

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