Is this a linear differential equation $\left(y^2-3x^2\right)\;dy +\left(2xy\right)\;dx=0$ I'm confused as to how I get started with this $\left(y^2-3x^2\right)\;dy +\left(2xy\right)\;dx=0$
Is it a linear differential equation?  Appreciate any hints that you can provide.
 A: No it's not a Linear ODE.
Linear ODE is of the Form below:
$$\sum_{i=1}^n P_i(x)y^{(n)}(x)+P_0(x)y+q(x)=0$$
where $y^{(n)}(x)$ is the n'th derivative of y with respect to x.
But there appear the term $y^2$ which is not linear.
this equation is Homogeneous Differential Equation of the first order. 
differential equation of first order of the form $M(x,y)dx+n(x,y)dy=0$ is called complete if $$\frac{\partial M}{\partial x}=\frac{\partial N}{\partial x}$$ but here it's not the case.
However we can make some equation of this form complete with a coefficient called Integral Factor.
The Mechanism is like Below:
$$\mu(x,y)M(x,y)dx+\mu(x,y)n(x,y)dy=0$$ 
now if we have 
$$\frac{\partial (\mu M)}{\partial x}=\frac{\partial (\mu N)}{\partial x}$$
then we can solve it. now finding this so called $\mu$.
well solving it using integral factor is almost impossible because Integral Factor is'nt a function of just one of variable $x$ or $y$, thus finding it is as hard as solving the main equation. instead consider this inventive way:
$$y'=\frac{2xy}{3x^2-y^2}$$
after some easy manipulation we get above equation. now we can write:
$$y'=\frac{2\frac{y}{x}}{3-(\frac{y}{x})^2}$$
now we introduce this new variable $\frac{y}{x}=h \Rightarrow y'=h'x+h$ so now we get this: 
$$h'x+h=\frac{2h}{3-h^2} \Rightarrow h'x=\frac{h(h+1)(h-1)}{3-h^2}$$ 
then we can solve it easily cause now the variables are separable. then with Partial Fraction Expansion and Integrating we get:
$$dh \times (\frac{-3}{h}+\frac{1}{h+1}+\frac{1}{h+1})=\frac{dx}{x}$$
$\Rightarrow $ln$(h-1)-3$ln$(h)+$ln$(h+1)=$ln$x+c \Rightarrow \frac{h^2-1}{h^3}=Cx \Rightarrow Cxh^3-h^2+1=0$
this is a cubic equation and has three solution. but unfortunately solving this equation to get solutions with Cardano's formula is quite complicated. but on bright side we still have the implicit formula for $y(x)$:
$$Cx(\frac{y}{x})^3-(\frac{y}{x})^2+1=0$$
