Given a general region, find the double integral bounded between $y = x$ and $y=3x-x^2 $ $$ J  = \iint_R (x^2-xy)\,dx \,dy, $$
Suppose region R is bounded between $y = x$ and $y=3x-x^2 $
My attempt using vertical integration:
$$ \int^{x=2}_{x=0} \int^{y=3x-x^2}_{y=x} \left({x^2-xy}\right)dy\ dx$$
$$\int^2_0 \left[x^2y-x\frac{y^2}{2}\right]^{3x-x^2}_{x}\, dx$$
$$\int^2_0 \frac{-x^5+4x^4-4x^3}{2} \,dx $$
$$\boxed{J = -\frac{8}{15}}$$

My attempt using horizontal integration :
$$ \int^{y=2}_{y=0} \int^{x=y}_{x=3\,\pm \sqrt{9-y}} \left({x^2-xy}\right)dx\ dy$$
For $ x = 3+\sqrt{9-y}$
$$  \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3+\sqrt{9-y} }\,dy$$
For $ x = 3-\sqrt{9-y}$
$$  \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3-\sqrt{9-y} }\,dy$$

My doubts :
1.) How do I set my limit of integration for horizontal integration, if there is $\pm$ to be considered ?
2.) the answer as negative what does that imply in questions related to double integrals?
Could you guys please help

Your first approach is correct, but your second approach is not. The possible values of $y$ lie in the interval $\left[0,\frac94\right]$, because $\frac94$ is the maximum of $3x-x^2$ when $x\in[0,2]$. When $y\in\left[0,\frac94\right]$, then (see the picture below):

*

*if $y\in[0,2]$, then the possible values of $x$ lie in $\left[\frac{3-\sqrt{9-4y}}2,y\right]$;

*if $y\in\left[2,\frac94\right]$, then the possible values of $x$ lie in $\left[\frac{3-\sqrt{9-4y}}2,\frac{3+\sqrt{9-4y}}2\right]$.

So, the answer is\begin{multline}\int_0^2\int_{\frac{3-\sqrt{9-4y}}2}^yx^2-xy\,\mathrm dy\,\mathrm dx+\int_2^{9/4}\int_{\frac{3-\sqrt{9-4y}}2}^{\frac{3+\sqrt{9-4y}}2}x^2-xy\,\mathrm dy\,\mathrm dx=\\=\int_0^2\frac{1}{12} \left(-2 y^3-6 y^2+\left(45-11 \sqrt{9-4 y}\right)  y+18 \left(\sqrt{9-4 y}-3\right)\right)\,\mathrm dy+\\+\int_2^{9/4}\frac{1}{6}(18-11 y)\sqrt{9-4 y}\,\mathrm dy=-\frac{47}{120}-\frac{17}{120}=-\frac8{15}.\end{multline}

