If ex+y+ey-x=1, then the value of y"-(y')2  is

ex+y+ey−x=1​⇒ex+e−x =e−y​      Differentiating w.r.t. x, we get​ex−e−x =−e−ydydx​Again differentiating w.r.t. x, we get​ex+e−x =−e−yd2ydx2+e−ydydx2 ​d2ydx2−dydx2                    =−eyex+e−x =−1