Answer
$y = g\left( x \right) = \sqrt {1 + \ln \left( {{{\sec }^2}x} \right)} $
Please see the figure attached.
Work Step by Step
We are given $f\left( {x,y} \right) = y\sin x$. The gradient is
$\nabla f = \left( {y\cos x,\sin x} \right)$
Let the curve $y = g\left( x \right)$ pass through $\left( {0,1} \right)$ that crosses each level curve of $f\left( {x,y} \right) = y\sin x$ at a right angle.
Since the gradient $\nabla f$ is normal to the level curves, the path parametrized by the form ${\bf{r}}\left( t \right) = \left( {t,g\left( t \right)} \right)$ passing through $\left( {0,1} \right)$ follow the gradient of $f\left( {x,y} \right) = y\sin x$. Then, the tangent vector ${\bf{r}}'\left( t \right)$ points in the direction of $\nabla f$ for all $t$. That is, ${\bf{r}}'\left( t \right) = k\left( t \right)\nabla {f_{{\bf{r}}\left( t \right)}}$ for some positive scalar function $k\left( t \right)$.
So, the equation ${\bf{r}}'\left( t \right) = k\left( t \right)\nabla {f_{{\bf{r}}\left( t \right)}}$ becomes
$\left( {1,g'\left( t \right)} \right) = k\left( t \right)\left( {g\left( t \right)\cos t,\sin t} \right)$
In component forms, we get
$1 = k\left( t \right)g\left( t \right)\cos t$,
$g'\left( t \right) = k\left( t \right)\sin \left( t \right)$
From the first equation we obtain $k\left( t \right) = \frac{1}{{g\left( t \right)\cos t}}$.
Substituting it in the second equation above gives
$g'\left( t \right) = \frac{{\tan t}}{{g\left( t \right)}}$
Since $g'\left( t \right) = \frac{{dg}}{{dt}}$, so
$gdg = \tan tdt$
Integrating both sides gives
$\smallint g{\rm{d}}g = \smallint \tan t{\rm{d}}t$
Recall from Example 6 of Section 8.2:
$\smallint \tan t{\rm{d}}t = \ln \left| {\sec t} \right| + constant$
So,
$\frac{1}{2}{g^2} = \ln \left| {\sec t} \right| + \ln c$,
where $c$ is integration constant.
${g^2} = \ln \left( {{c^2}{{\sec }^2}t} \right)$
Thus, the solution is $g\left( t \right) = \pm \sqrt {\ln \left( {{c_1}{{\sec }^2}t} \right)} $, where ${c_1} = {c^2}$ is a constant to be determined.
Since the path ${\bf{r}}\left( t \right) = \left( {t,g\left( t \right)} \right)$ passes through $\left( {0,1} \right)$ at $t=0$, we choose the positive solution $g\left( t \right) = \sqrt {\ln \left( {{c_1}{{\sec }^2}t} \right)} $ and get
${\bf{r}}\left( 0 \right) = \left( {0,\sqrt {\ln {c_1}} } \right)) = \left( {0,1} \right)$
So, ${c_1} = {\rm{e}}$. Thus, the path is ${\bf{r}}\left( t \right) = \left( {t,\sqrt {\ln \left( {{\rm{e}}{{\sec }^2}t} \right)} } \right)$
${\bf{r}}\left( t \right) = \left( {t,\sqrt {\ln \left( {{\rm{e}}{{\sec }^2}t} \right)} } \right)$
Since $\ln {\rm{e}} = 1$, thus
${\bf{r}}\left( t \right) = \left( {t,\sqrt {1 + \ln \left( {{{\sec }^2}t} \right)} } \right)$
Hence, we obtain: $y = g\left( x \right) = \sqrt {1 + \ln \left( {{{\sec }^2}x} \right)} $
