Step 1: Rearrange the differential equation.
Given:
\[
(1+y^2)\,dx-xy\,dy=0
\]
Rearranging,
\[
(1+y^2)\,dx=xy\,dy
\]
Therefore,
\[
\frac{dx}{x}=\frac{y}{1+y^2}\,dy
\]
Step 2: Integrate both sides.
Integrating,
\[
\int \frac{1}{x}\,dx
=
\int \frac{y}{1+y^2}\,dy
\]
\[
\log x
=
\frac12\log(1+y^2)+C
\]
Multiply by \(2\):
\[
2\log x=\log(1+y^2)+C
\]
Using logarithm properties:
\[
\log x^2=\log(1+y^2)+C
\]
Hence,
\[
x^2=C(1+y^2)
\]
Step 3: Use the initial condition.
Given:
\[
y(1)=0
\]
Substitute:
\[
1=C(1+0)
\]
\[
C=1
\]
Therefore,
\[
x^2=1+y^2
\]
or
\[
x^2-y^2=1
\]
Step 4: Identify the conic.
The equation
\[
x^2-y^2=1
\]
represents a hyperbola.
Comparing with
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,
\]
we get
\[
a^2=1,\quad b^2=1
\]
Eccentricity of hyperbola is
\[
e=\sqrt{1+\frac{b^2}{a^2}}
\]
So,
\[
e=\sqrt{1+\frac11}
\]
\[
e=\sqrt{2}
\]
Step 5: Final conclusion.
Therefore, the eccentricity is
\[
\boxed{\sqrt{2}}
\]