If $\ln \left( {(e - 1){e^{xy}} + {x^2}} \right) = {x^2} + {y^2}$,then $\left. {\frac{{dy}}{{dx}}} \right|_{(1,0)}$ is

Let $C$ be the curve $y^3 - 3xy + 2 = 0$. If $H$ and $V$ are the sets of points on the curve $C$ where the tangent to the curve is horizontal and vertical respectively,then

The curve $3y^2 = 2ax^2 + 6b$ passes through the point $P(3, -1)$ and the gradient of the curve at $P$ is $-1$. Then the values of $a$ and $b$ are:

For the curve $\sqrt{x} + \sqrt{y} = 1$,the value of $\frac{dy}{dx}$ at the point $\left( \frac{1}{4}, \frac{1}{4} \right)$ is:

If $y \sec x + \tan x + x^2 y = 0$,then $\frac{dy}{dx} =$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable odd function satisfying $f^{\prime}(x) \geq 0$,$f^{\prime\prime}(x) = f(x)$,$f(0) = 0$,and $f^{\prime}(0) = 3$. Then $9f(\log_e 3)$ is equal to . . . . . . .