if sqrt(y+x)+ sqrt(y-x)=c, show that dy/dx=y/x-sqrt(y²/x²-1).

the given equation is : $\sqrt{y+x}+\sqrt{y-x}=c$ ..........(1)
differentiating wrt x:
​$$\begin{gathered} \frac{1}{2\sqrt{y+x}}.\left[\frac{dy}{dx}+1\right]+\frac{1}{2\sqrt{y-x}}.\left[\frac{dy}{dx}-1\right]=0 \\ \left[\frac{1}{\sqrt{y+x}}+\frac{1}{\sqrt{y-x}}\right].\frac{dy}{dx}+\left[\frac{1}{\sqrt{y+x}}-\frac{1}{\sqrt{y-x}}\right]=0 [multipliyng eq by 2] \end{gathered}$$
$$\begin{gathered} \left[\frac{\sqrt{y-x}+\sqrt{y+x}}{\sqrt{(y-x)(y+x)}}\right].\frac{dy}{dx}+\left[\frac{\sqrt{y-x}-\sqrt{y+x}}{\sqrt{(y-x)(y+x)}}\right]=0 \\ \left[\sqrt{y-x}+\sqrt{y+x}\right].\frac{dy}{dx}+\left[\sqrt{y-x}-\sqrt{y+x}\right]=0 [multiplying whole eq. by \sqrt{(y-x)(y+x)}] \end{gathered}$$
$$\begin{gathered} \frac{dy}{dx}=-\frac{\sqrt{y-x}-\sqrt{y+x}}{\sqrt{y-x}+\sqrt{y+x}} \\ \frac{dy}{dx}=-\frac{\sqrt{y-x}-\sqrt{y+x}}{\sqrt{y-x}+\sqrt{y+x}}*\frac{\sqrt{y-x}-\sqrt{y+x}}{\sqrt{y-x}-\sqrt{y+x}} [rationali\sin g the denominator] \\ \frac{dy}{dx}=-\frac{{\left(\sqrt{y-x}-\sqrt{y+x}\right)}^2}{(y-x)-(y+x)} \end{gathered}$$
$$\begin{gathered} \frac{dy}{dx}=-\frac{y-x+y+x-2.\sqrt{y-x}.\sqrt{y+x}}{-2x} \\ \frac{dy}{dx}=\frac{2y-2\sqrt{(y-x)(y+x)}}{2x}=\frac{y-\sqrt{(y-x)(y+x)}}{x} \\ \frac{dy}{dx}=\frac{y}{x}-\frac{\sqrt{y^2-x^2}}{x} \\ \frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1} \end{gathered}$$
which is the required result.
hope this helps you