What is the order of the differential equation dx dy +∫ ydx=x ^3 ?

$$\begin{aligned} & \frac{d x}{d y}+\int y \cdot d x=x^3 \Rightarrow \int y \cdot d x=x^3-\frac{d x}{d y} \\ & \Rightarrow 1+\frac{d y}{d x}\left(\int y \cdot d x\right)=x^3 \cdot \frac{d y}{d x} \end{aligned}$$ Differentiate both sides w.e.t. $x$$\begin{aligned} & \Rightarrow 0+\frac{d y}{d x}(y)+\left(\int y \cdot d x\right)\left(\frac{d^2 y}{d x^2}\right)=x^3 \cdot \frac{d^2 y}{d x^2}+\frac{d y}{d x}\left(2 x^2\right) \\ & \Rightarrow y \cdot \frac{d y}{d x}+\frac{d^2 y}{d x^2}\left[x^3-\frac{d x}{d y}\right]=x^3 \cdot \frac{d^2 y}{d x^2}+2 x^2 \frac{d y}{d x} \\ & \Rightarrow y \frac{d y}{d x}+x^3 \frac{d^2 y}{d x^2}-\left(\frac{d x}{d y}\right)\left(\frac{d^2 y}{d x^2}\right)=x^3 \frac{d^2 y}{d x^2}+2 x^2 \frac{d y}{d x} \\ & \Rightarrow y \frac{d y}{d x}-\frac{d x}{d y} \cdot \frac{d^2 y}{d x^2}=2 x^2 \cdot \frac{d y}{d x} \end{aligned}$Multiplying both side by $\frac{d y}{d x}\begin{aligned} & y\left(\frac{d y}{d x}\right)^2-\frac{d^2 y}{d x^2}=2 x^2\left(\frac{d y}{d x}\right)^2 \\ & \Rightarrow \frac{d^2 y}{d x^2}+\left(2 x^2-y\right)\left(\frac{d y}{d x}\right)^2=0 \\ & \text{ Order }=2, \text{ degree }=1 \end{aligned}$$$