Solution: We want to solve the differential equation \[ \frac{\mathrm{d} y}{\mathrm{d} x} + y\tan x = y^3 \sec^4 x. \] This is the Bernoulli's differential equation. It is in the form of \[ y' + p\left( x \right)y = q\left( x \right){y^n}, \] where $p(x)$ and $q\left( x \right) $ are continuous functions. In order to solve the differential equation we will follow the following steps.

1. Divide both sides of the equation by $y^n$: \begin{equation}\label{eq:17Mar2023-1} \frac{y^\prime }{y^n} + \frac{p(x)}{y^{n-1}} = q(x). \end{equation}
2. Now substitute $z=\frac{1}{y^{n-1}}$ which implies \[ \frac{\mathrm{d}z}{\mathrm{d} x} = \frac{-(n-1)}{y^n} \frac{\mathrm{d} y}{\mathrm{d} x} \implies \frac{y^\prime }{y^n} = \frac{z^\prime }{1-n}. \]
3. Substitute the above in \eqref{eq:17Mar2023-1} to get \begin{equation}\label{eq:17Mar2023-2} \frac{z^\prime }{1-n} + z p(x) = q(x). \end{equation}
4. The equation \eqref{eq:17Mar2023-2} is a linear differential equation which can be solved by multiplying the differential equation by an integrating factor.

We will now solve the given differential equation. We have \[ p(x) = \tan x,~q(x) = \sec^4 x,~\text{and } n = 3. \] Following the above steps we have \begin{equation}\label{eq:17Mar2023-3} z^\prime -2 z \tan x = -2\sec ^4x. \end{equation} As explained above, we can use the integrating factor method to solve the linear differential equation. Integrating factor will be \begin{equation}\label{eq:17Mar2023-4} \mu(x) = e^{\int -2 \tan x ~\mathrm{d} x} = e^{-2 \ln \vert \sec x \vert } = \frac{1}{\sec ^2x}. \end{equation}

Multiplying \eqref{eq:17Mar2023-3} by the integrating factor. \begin{equation*} \frac{z^\prime }{\sec ^2x} - \frac{2z \tan x}{\sec ^2x} = -2\sec ^2x. \end{equation*} Now observe that \begin{align*} & \frac{z^\prime }{\sec ^2x} - \frac{2z \tan x}{\sec ^2x} = -2\sec ^2x \\ \implies & \left( \frac{z}{\sec ^2x} \right)^\prime = -2\sec ^2x \\ \implies & \int \left( \frac{z}{\sec ^2x} \right)^\prime = \int -2\sec ^2x~\mathrm{d} x \\ \implies & \frac{z}{\sec ^2x} = c - 2\tan x \\ \implies & z = (c-2\tan x)\sec ^2x. \end{align*} So we have \begin{align*} z(x) = \frac{1}{y^2} \implies y^2 = \frac{1}{(c-2\tan x)\sec ^2x}. \end{align*} Therefore, \[ \textcolor{blue}{\boxed{y(x) = \frac{1}{\sqrt{(c-2\tan x)\sec ^2x} } \text{ or } y(x) = -\frac{1}{\sqrt{(c-2\tan x)\sec ^2x} } }} \]