Many thanks Udo,for helping me and patience to explain,thanks again

Hi Luis, just take a step back and consider $x$ and $y$ as independent variables (i.e. not $y=y(x)$). Then search for a new function $f=f(x,y)$, the so-called potential function with the property that the total differential of $f$ gives the differential equation back: $\begin{equation}df=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy = (5x+4y)dx+(4x-8y^3)dy\end{equation}=0$. To do this one solves a partial differential equation system for $f$ in $x$ and $y$

In[1]:= DSolve[{D[f[x, y], x] == 5 x + 4 y, D[f[x , y], y] == 4 x - 8 y^3}, f, {x, y}]
Out[1]= {{f -> Function[{x, y}, (5 x^2)/2 + 4 x y - 2 y^4 + C[1]]}}

which gives clearly your lecturer's solution.

The constant C[1] does not matter, because it disappears under the total differential of $f$.

My Spanish is zero, but the ordinary differential equation is exact. Your lecturer did not solve for $y(x)$ but for the potential function $f(x,y)$. The ordinary differential equation is the total differential of $f$:

In[4]:= (D[#, x] DifferentialD[x] + D[#, y] DifferentialD[y]) &[5 x^2/2 + 4 x y - 2 y^4]
Out[4]= (5 x + 4 y) \[DifferentialD]x + (4 x - 8 y^3) \[DifferentialD]y

Please note that differentiation for $x$ did not produce terms $\frac{dy}{dx}$. By the way, you committed a mistake citing it as

(5x + 4y) dx +(4x + 8y^3) dy =0

(5x + 4y) dx +(4x + 8y^3) dy =0

If you divide by $dy$ it becomes $(5 x(y) + 4 y) \frac{dx}{dy}+(4 x(y)+ 8 y^3) = 0$. So the dependent variable is $x$ and the independent variable is $y$, hence

ode = (5 x[y] + 4 y) x'[y] + (4 x[y] + 8 y^3) == 0;
DSolve[ode, x[y], y]

If $y$ was the dependent variable and $x$ is the independent variable, then dividing by $dx$ gives the ode $(5 x + 4 y(x))+(4 x+ 8 y^3(x)) \frac{dy}{dx} = 0$ and now you write

ode = (5 x + 4 y[x]) + (4 x + 8 y[x]^3) y'[x] == 0;
DSolve[ode, y[x], x]