= Solution
The <homogeneous first-order differential equation> suggests $v=y/x$, so $y'=v+xv'$. Wherever the original equation is defined,
$$
xv'=-\tan v,\qquad\frac{d}{dx}[x\sin v]=\sin v+x\cos v\,v'=0.
$$
Thus the implicit solution is $x\sin(y/x)=C$. The printed initial values force $C=x_0=2/\pi$.
\b[There is no classical solution through the printed initial point.] There $y/x=\pi/2$, so the tangent on the right is undefined. The implicit relation nevertheless has two neighboring real branches on $x>x_0$ which approach that point continuously:
$$
\boxed{y_1(x)=x\arcsin(x_0/x),\qquad y_2(x)=x[\pi-\arcsin(x_0/x)],\quad x_0=2/\pi.}
$$
Both tend to $y=1$ as $x\downarrow x_0$, but their <derivatives> become unbounded with opposite signs. They solve the equation on the open interval $x>x_0$, not at its endpoint. Other inverse-sine periods cannot approach the specified initial value. This is an <endpoint obstruction in a homogeneous differential equation>. Constant-ratio branches $y=k\pi x$ also solve the general equation where $x\ne0$, but do not meet the initial data.
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