Step 1: Separate variables:
\(2x\dfrac{dy}{dx}=y+3\), so \(\dfrac{dy}{y+3}=\dfrac{dx}{2x}\).
Step 2: Integrate:
\(\ln|y+3|=\dfrac12\ln|x|+\ln C\), so \(\ln|y+3|=\ln(C\sqrt{|x|})\).
Step 3: Simplify:
Squaring both sides: \((y+3)^2=C'x\). This is of the form \(Y^2=4aX\), a parabola.
Step 4: Why the other options are wrong.
A straight line would be of degree 1 in \(x\) and \(y\). A circle or ellipse needs squares of both \(x\) and \(y\), but here only \(y\) is squared.
Final Answer:
The solution is a family of parabolas.
\[ \boxed{\text{(C) }\text{parabola}} \]