Since $\frac{x^{2}+2}{x+3}$ and $\frac{x\left(x+3\right)}{x+3}$ have the same denominator, subtract them by subtracting their numerators.
$$\frac{x^{2}+2-x\left(x+3\right)}{x+3}>0$$
Do the multiplications in $x^{2}+2-x\left(x+3\right)$.
$$\frac{x^{2}+2-x^{2}-3x}{x+3}>0$$
Combine like terms in $x^{2}+2-x^{2}-3x$.
$$\frac{2-3x}{x+3}>0$$
For the quotient to be positive, $2-3x$ and $x+3$ have to be both negative or both positive. Consider the case when $2-3x$ and $x+3$ are both negative.
$$2-3x<0$$ $$x+3<0$$
This is false for any $x$.
$$x\in \emptyset$$
Consider the case when $2-3x$ and $x+3$ are both positive.
$$x+3>0$$ $$2-3x>0$$
The solution satisfying both inequalities is $x\in \left(-3,\frac{2}{3}\right)$.
$$x\in \left(-3,\frac{2}{3}\right)$$
The final solution is the union of the obtained solutions.
