I'm guessing the equation should read something like
[tex]7(x^2+y^2) \,dx + 5xy \, dy = 0[/tex]
or possibly with minus signs in place of +.
Multiply both sides by [tex]\frac1{x^2}[/tex] to get
[tex]7\left(1 + \dfrac{y^2}{x^2}\right) \, dx + \dfrac{5y}x \, dy = 0[/tex]
Now substitute
[tex]v = \dfrac yx \implies y = xv \implies dy = x\,dv + v\,dx[/tex]
to transform the equation to
[tex]7(1+v^2) \, dx + 5v (x\,dv + v\,dx) = 0[/tex]
which simplifies to
[tex](7 + 12v^2) \, dx + 5xv\,dv = 0[/tex]
The ODE is now separable.
[tex]\dfrac{5v}{7+12v^2} \, dv = -\dfrac{dx}x[/tex]
Integrate both sides. On the left, substitute
[tex]w = 7+12v^2 \implies dw = 24v\, dv[/tex]
[tex]\displaystyle \int \frac{5v}{7+12v^2} \, dv = -\int \frac{dx}x[/tex]
[tex]\displaystyle \dfrac5{24} \int \frac{dw}w = -\int \frac{dx}x[/tex]
[tex]\dfrac5{24} \ln|w| = -\ln|x| + C[/tex]
Solve for [tex]w[/tex].
[tex]\ln\left|w^{5/24}\right| = \ln\left|\dfrac1x\right| + C[/tex]
[tex]\exp\left(\ln\left|w^{5/24}\right|\right) = \exp\left(\ln\left|\dfrac1x\right| + C\right)[/tex]
[tex]w^{5/24} = \dfrac Cx[/tex]
Put this back in terms of [tex]v[/tex].
[tex](7+12v^2)^{5/24} = \dfrac Cx[/tex]
Put this back in terms of [tex]y[/tex].
[tex]\left(7+12\dfrac{y^2}{x^2}\right)^{5/24} = \dfrac Cx[/tex]
Solve for [tex]y[/tex].
[tex]7+12\dfrac{y^2}{x^2} = \dfrac C{x^{24/5}}[/tex]
[tex]\dfrac{y^2}{x^2} = \dfrac C{x^{24/5}} - \dfrac7{12}[/tex]
[tex]y^2= \dfrac C{x^{14/5}} - \dfrac{7x^2}{12}[/tex]
[tex]y = \pm \sqrt{\dfrac C{x^{14/5}} - \dfrac{7x^2}{12}}[/tex]
