$u = 4x = \frac{{\partial \psi }}{{\partial y}}$ 

$\Rightarrow \partial \psi  = 4x\partial y$

$\Rightarrow \int {\partial \psi  = \int {4x\partial y} }$

$\Rightarrow \psi  = 4xy + f\left( x \right)$

                                      $v =  – 4y$ = -$\frac{{ \partial \psi }}{{\partial x}}$

$\Rightarrow \frac{{\partial \psi }}{{\partial x}} = 4y$

$\Rightarrow \partial \psi  = 4y\partial x$

$\Rightarrow \int {\partial \psi  = \int {4y\partial x} }$

$\Rightarrow \psi  = 4xy + f\left( y \right)$

On comparing these two equations for $\psi$, stream functions is $\psi  = 4xy + {\rm{constant}}$.

Also                               $u = 4x = \frac{{\partial \phi }}{{\partial x}}$

$\Rightarrow 4x\partial x = \partial \phi$

$\Rightarrow \int {\partial \phi  = \int {4x\partial x} }$

$\Rightarrow \phi  = 4{x^2} + f\left( y \right)$

                                      $v =  – 4y = \frac{{\partial \phi }}{{\partial y}}$

$\Rightarrow \frac{{\partial \phi }}{{\partial y}} =  – 4y$

$\Rightarrow \int {\partial \phi  = \int { – 4y\partial y} }$

$\Rightarrow \phi  =  – 4{y^2} + f\left( x \right)$

On comparing the above two equations, $f\left( y \right) =  – 4{y^2}$ and $f(x) = 4{x^2}$, velocity potential is $\phi  = 4\left( {{x^2} – {y^2}} \right)$.

Now

$\psi  = 4xy + {\rm{constant}}$, differentiating with respect to x, holding $\psi  = {\rm{constant}}$

$0 = 4x\frac{{dy}}{{dx}} + 4y$

$\Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\psi  = {\rm{constant}}}} =  – \frac{y}{x}$

Differentiating $\phi  = 4\left( {{x^2} – {y^2}} \right)$ with respect to x, holding $\phi  = {\rm{constant}}$ 

$0 = 2 \times 4x – 2 \times 4 \times y\frac{{dy}}{{dx}}$

$\Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\phi  = {\rm{constant}}}} = \frac{x}{y}$