A streamline in a fluid flow

Velocity field is $V=\left ( 3yt\hat{i}+5x\hat{j} \right )$
In a streamline flow, velocity is tangent to the flow. $V$ and $dr$ is in the same direction. Therefore,
$$V\times dr=0$$ Since, two vectors are in the same direction, their cross product is $0$.
$$V\times dr = 0$$ $$\Rightarrow \left ( 3yt\hat{i}+5x\hat{j} \right )\times \left ( dx\hat{i}+dy\hat{j} \right )=0$$
$$\Rightarrow \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 3yt& 5x &0 \\ dx&dy & 0 \end{vmatrix}=0$$
$$\Rightarrow \hat{k}\left ( 3ytdy-5xdx \right )=0$$
At, time $t = 3$
$$\Rightarrow 9y\left ( 3 \right )dy -5xdx=0$$ $$\Rightarrow 9ydy -5xdx=0$$ $$\Rightarrow 9ydy=5xdx$$
On integrating
$$\Rightarrow \int 9ydy=\int 5xdx$$
$$\Rightarrow 9\int ydy=5\int xdx$$
$$\Rightarrow 9\frac{y^{2}}{2}+c_{1}=\frac{5x^{2}}{2}+c_{2}$$
$$\Rightarrow \frac{9y^{2}+2c_{1}}{2}=\frac{5x^{2}+2c_{2}}{2}$$
$$\Rightarrow 9y^{2}+2c_{1}=5x^{2}+2c_{2}$$
$$\Rightarrow 9y^{2}-5x^{2}=2c_{2}-2c_{1}$$
$$\Rightarrow 9y^{2}-5x^{2}=C$$

At, point $(5,3)$, since,
$$9y^{2}-5x^{2}=C$$ $$\Rightarrow C= 9\left ( 3 \right )^{2}-5\left ( 5 \right )^{2}$$ $$\Rightarrow C=\left ( 9\times 9 \right )-\left ( 5\times 25 \right )$$ $$\Rightarrow C=81-125=-44$$ Therefore, the equation of the streamline is,
$$9y^{2}-5x^{2}=-44$$ $$\Rightarrow 9y^{2}-5x^{2}+44=0$$