Using implicit differentiation, find $\dfrac {dy}{dx}$ $$x^3y^6=(x+y)^9$$
My Attempt: $$x^3y^6=(x+y)^9$$ Differentiating both sides with respect to $x$, we get: $$\dfrac {d}{dx} (x^3y^6)=\dfrac {d}{dx} ((x+y)^9)$$ $$x^3.\dfrac {d}{dx} y^6 + y^6. \dfrac {d}{dx} x^3=9(x+y)^8 \dfrac {d}{dx} (x+y)$$ $$6x^3y^5.\dfrac {dy}{dx} + 3x^2y^6 = 9(x+y)^8 (1+\dfrac {dy}{dx})$$ $$\dfrac {dy}{dx}=\dfrac {9(x+y)^8-3x^2y}{6x^3y^5-9(x+y)^8}$$
But the answer is $$\dfrac {dy}{dx}=\dfrac {y}{x}$$...
I fixed a typo in your question, and with that, your implicit differentiation is perfectly correct. However, it looks difficult (as in, I cannot see how) to reduce it from that to the expected simple answer.
However, following @Arthur's hint, if you're dealing only with the real domain it makes things a lot easier to take cube roots right at the outset. This simplifies things greatly.
$x^3y^6 = (x+y)^9$
For $x,y \in \mathbb{R}$,
$xy^2 = (x+y)^3$
Differentiating implictly wrt $x$,
$y^2 + 2yy'x = 3(x+y)^2(1+y')$
$y'(2yx-3(x+y)^2) = 3(x+y)^2 - y^2$
$\displaystyle y' = \frac{3(x+y)^2 - y^2}{2yx-3(x+y)^2}$
Use the initial relationship given on the RHS,
$\displaystyle y' = \frac{\frac{3(x+y)^3}{x+y} - y^2}{2yx-\frac{3(x+y)^3}{x+y}}$
$\displaystyle y' = \frac{\frac{3xy^2}{x+y} - y^2}{2yx-\frac{3xy^2}{x+y}}$
The rest is just algebra,
$\displaystyle y' = \frac{3xy^2 - y^2(x+y)}{2yx(x+y)-3xy^2}$
$\displaystyle y' = \frac{3xy^2 - xy^2 - y^3}{2yx^2 + 2y^2x-3xy^2}$
$\displaystyle y' = \frac{2xy^2 - y^3}{2yx^2 - xy^2}$
$\displaystyle y' = \frac{y^2(2x - y)}{yx(2x - y)}$
$\displaystyle y' = \frac yx$
as required, et voilà!
EDIT: And just to flesh out @lab's excellent hint involving logarithmic differentiation:
$3 \log x + 6\log y = 9\log(x+y)$
$\displaystyle \frac 3x + \frac 6y y'= \frac 9{x+y}(1+y')$
$\displaystyle y'(\frac 6y - \frac 9{x+y}) = \frac 9{x+y} - \frac 3x$
$\displaystyle y'(\frac{6x-3y}{y(x+y)}) = \frac{6x-3y}{x(x+y)}$
$\displaystyle y' = \frac yx$
A decidedly much quicker route to the answer.