Step 1: Rewrite the equation.
Cross-multiply the given equation $\frac{dy}{dx}=\frac{x}{y}$ to get $y\,dy-x\,dx=0$.
Step 2: Spot an exact differential.
Notice that $y\,dy=d\left(\frac{y^{2}}{2}\right)$ and $x\,dx=d\left(\frac{x^{2}}{2}\right)$. So the equation says $d\left(\frac{y^{2}-x^{2}}{2}\right)=0$.
Step 3: Conclude.
A quantity with zero differential is constant. So $y^{2}-x^{2}=c$ for some constant $c$.
Step 4: Identify the conic.
Compare with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Our curve has squares with opposite signs, which is the hyperbola. Taking different values of $c$ gives different hyperbolas, so we get a family.
Step 5: Check with a slope test.
On a circle centred at the origin, the slope is $-\frac{x}{y}$, which is the opposite of the given slope $\frac{x}{y}$. So circles are ruled out.
Final Answer:
Option 4 is correct. \[ \boxed{\text{hyperbolas}} \]