A curve is such that $\dfrac{d^2y}{dx^2}=3e^{2x}-e^{-x}$. The curve cuts the y-axis at P (0, 1) and has a gradient of 4 at P.

1. Integrate the second derivative to find the general expression for the first derivative.

$\frac{dy}{dx} = \int (3e^{2x} - e^{-x}) \, dx$
$\frac{dy}{dx} = \frac{3}{2}e^{2x} + e^{-x} + C$

2. Find the constant of integration using the given gradient at $x=0$.

At $P(0, 1)$, $x = 0$ and $\frac{dy}{dx} = 4$.
$4 = \frac{3}{2}e^{0} + e^{0} + C$
$4 = \frac{3}{2} + 1 + C$
$C = \frac{3}{2}$
$\frac{dy}{dx} = \frac{3}{2}e^{2x} + e^{-x} + \frac{3}{2}$

3. Show that the first derivative is never zero.

For a stationary point to exist, $\frac{dy}{dx} = 0$.
However, for all real $x$, $e^{2x} > 0$ and $e^{-x} > 0$.
Therefore, $\frac{3}{2}e^{2x} + e^{-x} + \frac{3}{2} > 0$ for all real $x$.
Since $\frac{dy}{dx} \neq 0$ for all $x$, the curve has no stationary points.

1. Find the expression for the gradient, $\frac{dy}{dx}$, by integrating $\frac{d^2y}{dx^2}$ and using the given gradient at $P$.

$\frac{dy}{dx} = \int (3e^{2x} - e^{-x}) \, dx$
$\frac{dy}{dx} = \frac{3}{2}e^{2x} + e^{-x} + c$
At $x = 0$, $\frac{dy}{dx} = 4$:
$\frac{3}{2}e^{0} + e^{0} + c = 4$
$\frac{3}{2} + 1 + c = 4 \implies c = \frac{3}{2}$
$\frac{dy}{dx} = \frac{3}{2}e^{2x} + e^{-x} + \frac{3}{2}$

2. Integrate $\frac{dy}{dx}$ to find the general equation for $y$.

$y = \int \left(\frac{3}{2}e^{2x} + e^{-x} + \frac{3}{2}\right) \, dx$
$y = \frac{3}{4}e^{2x} - e^{-x} + \frac{3}{2}x + d$

3. Use the coordinates of $P$ to find the constant of integration and state the final equation.

At $x = 0$, $y = 1$:
$\frac{3}{4}e^{0} - e^{0} + \frac{3}{2}(0) + d = 1$
$\frac{3}{4} - 1 + d = 1$
$-\frac{1}{4} + d = 1 \implies d = \frac{5}{4}$

Equation of the curve:
$y = \frac{3}{4}e^{2x} - e^{-x} + \frac{3}{2}x + \frac{5}{4}$