If $\sin(xy) + \cos(xy) = 0$,then $\frac{dy}{dx} = $

If $f(x)$ is an invertible and twice differentiable function satisfying $f'(x) = \int_{0}^{f(x)} f^{-1}(t) dt$ for all $x \in R$ and $f'(0) = 1$,then $f'(1)$ is equal to:

If $x^{2}+y^{2}=t+\frac{1}{t}$ and $x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$,then find $\frac{d y}{d x}$.

If $x{e^{xy}} = y + {\sin ^2}x$,then at $x = 0$,$\frac{dy}{dx} = $

If $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then $\frac{dy}{dx}=$

For the curve $C : (x^{2}+y^{2}-3)+(x^{2}-y^{2}-1)^{5}=0$,the value of $3y^{\prime}-y^{3}y^{\prime\prime}$ at the point $(\alpha, \alpha)$,where $\alpha > 0$,on $C$ is equal to: