Solving the differential equation $−3xdx+(x^2y+3y)dy=0$ 
Solve the following differential equation by finding an appropriate integrating factor
  $$ −3x\,\mathrm{d}x+(x^2y+3y)\,\mathrm{d}y=0$$

My attempt:
The equation is the same as: $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3x}{(x+3)y}$$
Make y=z^{1/2}
Then $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{2}z^{-1/2}\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{3x}{x^2+3}y^{-1}$$
replace y with $$y=z^{1/2}
$$\frac{1}{2}z^{-1/2}\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{3x}{x^2+3}z^{-1/2}$$
$$
z=3\ln(x^2+3)+C, \quad \text{ while } \quad y=z^{1/2}
$$
so
$$
y^2-3\ln(x^2+3)=c
$$
screenshot is here:


the assignment system said my attemp was wrong without giving the correct answer. Please let me know where I am wrong. thanks a lot!

thanks for the help guys. it turns to be a bug of the website.
 A: $-3xdx+(x^2+3y)dy=0$ is equivalent to $(x^2y+3y)dy=3xdx$
$$\frac{dy}{dx}=\frac{3x}{(x^2+3)y}$$
By letting $y=\sqrt z$, we get,
$$\frac{dy}{dx}=\frac{dz}{2\sqrt zdx}=\frac {3x}{(x^2+3)\sqrt z}$$
$${dz}=\frac{6x\cdot dx}{x^2+3}$$
Integrate both side, for the RHS we can use $u$-substitution with $u=x^2+3$.
$$y^2=z=3\ln(x^2+3)+C$$
$$y^2-3\ln(x^2+3)=C$$
There are $1\over 2$ that you forget to multiply with before you integrate $dz$.
A: The equation is separable, you can rewrite it as 
$$
yy'=\frac{3x}{x^2+3}.
$$
Integrating you get (via the substitution $v=x^2+3$), 
$$
\frac{y^2}2=\frac32\,\ln(x^2+3)+c.
$$
After multiplying by $2$ and renaming $c$ you get 
$$
y^2=3\ln(x^2+3)+c.
$$
So your answer is correct. The "integrating factor" thing means probably that they want you to force the equation to be exact by multiplying by an appropriate function. Not that it changes the answer. 
A: $$−3x\,\mathrm{d}x+(x^2y+3y)\,\mathrm{d}y=0$$
$$−3x\,\mathrm{d}x+(x^2+3)y\,\mathrm{d}y=0$$
$$−3\frac {x}{(x^2+3)}\,\mathrm{d}x+y\,\mathrm{d}y=0$$
$$−\frac 32\frac {2x}{(x^2+3)}\,\mathrm{d}x+\frac 12 {d}y^2=0$$
$$−\frac 32\,\mathrm{d}\ln |{(x^2+3)}|+\frac 12 {d}y^2=0$$
Integrate:
$$− 3\ln |{(x^2+3)}|+ y^2=C$$
Therefore:
$$\boxed {y^2− 3\ln {(x^2+3)}=C}$$
Your answer seems correct to me. Maybe the webpage is expecting another form for the solution.Did you try to type the absolute value for the $\ln$ function ? Maybe it's just a formatting question. Try:
$$− 3\ln |x^2+3|+ y^2=C$$
