$x$ and $y$ components of velocity are given by $u=cx$ and $v=-cy$.We need to calculate the stream function and velocity potential.Here

$$u=cx \\\frac{\partial \psi }{\partial y}=cx \\\Rightarrow \partial \psi=cx\partial y \\\Rightarrow \psi=cxy+f(x) \\v=-cy=-\frac{\partial \psi}{\partial x} \\\Rightarrow \psi=cxy+f(y)$$

Here f(x) and f(y) are constants:

$$\psi=cxy+const. \\u=cx=\frac{d\phi}{dx} \\\Rightarrow d\phi=cxdx \\\Rightarrow \phi=cx^{2}+f(y) \\v=-cy=\frac{\partial \phi}{\partial y}:\phi=-cy^{2}+f(x)$$

Therefore $$f(y)=-cy^{2}\;and \;f(x)=cx^{2}$$

and $$\phi=c\left ( x^{2}-y^{2} \right ) \\\psi=cxy+const.$$

Differentiating the above equation of $\psi$ with respect to ‘$x$’, holding $\psi=constant$

$$\frac{\partial \psi}{\partial x}=cx\frac{dy}{dx}+cy$$

or $$\left ( \frac{dy}{dx} \right )_{\psi=constant}=\left ( \frac{-y}{x} \right )$$

Differentiating the above equation of $\phi$ with respect to ‘$x$’,holding $\phi=constant$

$$0=2cx-2cy\left ( \frac{dy}{dx} \right )$$

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

On comparing the above equations we get $$\left ( \frac{dy}{dx} \right )_{\psi=constant}=\frac{-1}{\left (\frac{dy}{dx} \right )_{\phi=constant}}$$

Therefore the lines of constant $\phi$ are perpendicular to the lines of constant $\psi$.