The differential equation $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\mathit{F}\left(\mathit{x},\mathit{y}\right)$ is homogenous differential equation if $\mathit{F}\left(\mathit{\lambda }\mathit{x},\mathit{\lambda }\mathit{y}\right)=\mathit{F}\left(\mathit{x},\mathit{y}\right)$.
Examining the above options, (B), (C) and (D) are homogeneous differential equation as below.
(B) $\mathit{F}\left(\mathit{\lambda }\mathit{x},\mathit{\lambda }\mathit{y}\right)=\frac{\mathit{\lambda }\mathit{y}}{\mathit{\lambda }\mathit{x}}=\frac{\mathit{y}}{\mathit{x}}=\mathit{F}\left(\mathit{x},\mathit{y}\right)$
(C) $\mathit{F}\left(\mathit{\lambda }\mathit{x},\mathit{\lambda }\mathit{y}\right)=\frac{{\left(\mathit{\lambda }\mathit{x}\right)}^2+\left(\mathit{\lambda }{\mathit{y}}^2\right)}{\mathit{\lambda }\mathit{x}\mathit{\lambda }\mathit{y}}=\frac{{\mathit{\lambda }}^2{\mathit{x}}^2+{\mathit{\lambda }}^2{\mathit{y}}^2}{{\mathit{\lambda }}^2\mathit{x}\mathit{y}}=\frac{{\mathit{x}}^2+{\mathit{y}}^2}{\mathit{x}\mathit{y}}=\mathit{F}\left(\mathit{x},\mathit{y}\right)$
(D) $\mathit{F}\left(\mathit{\lambda }\mathit{x},\mathit{\lambda }\mathit{y}\right)={\cos }^2\left(\frac{\mathit{\lambda }\mathit{y}}{\mathit{\lambda }\mathit{x}}\right)={\cos }^2\left(\frac{\mathit{y}}{\mathit{x}}\right)=\mathit{F}\left(\mathit{x},\mathit{y}\right)$
Now we examine (A) as below:
(A) $\mathit{F}\left(\mathit{\lambda }\mathit{x},\mathit{\lambda }\mathit{y}\right)=\cos \mathit{\lambda }\mathit{x}-\sin \left(\frac{\mathit{\lambda }\mathit{y}}{\mathit{\lambda }\mathit{x}}\right)=\cos \mathit{\lambda }\mathit{x}-\sin \left(\frac{\mathit{y}}{\mathit{x}}\right)\mathrm{ne}\mathit{F}\left(\mathit{x},\mathit{y}\right)$, so (A) is not a homogeneous differential equation.