I love being inspired! 7

This would definitely reduce some calculations. Substituting y=mx in the second equation, where m=cot $\theta$ and cross multiplication gives(along with cancellation of $x^4$ in denominators): $\sin^8\theta+\cos^8\theta=194\sin^4\theta \cos^4\theta$ $\Rightarrow (\sin^4\theta+cos^4\theta)^2=196\sin^4\theta \cos^4\theta$ $\Rightarrow \sin^4\theta+\cos^4\theta=14\sin^2\theta \cos^2\theta$ $\Rightarrow \sin^2\theta+\cos^2\theta=4\sin\theta \cos\theta$ $\Rightarrow \tan\theta+\cot\theta=4$ Hence $\frac{x}{y}+\frac{y}{x}=\tan\theta+\cot\theta \space =\color{#20A900}{4}\\$ $\color{#3D99F6}{\text {Beauty of algebra and trigonometry!!}}$

This is a very very beatiful problem.

So let us see what we have got,

$\frac{\sin\theta}{x}=\frac{\cos\theta}{y}$

And,

$\frac{\cos^{4}\theta}{x^{4}}+\frac{\sin^{4}\theta}{y^{4}}=\frac{97\sin 2\theta}{x^{3}y+y^{3}x}$

Let us just simplify the second equation,

$\frac{\cos^{4}\theta}{x^{4}}+\frac{\sin^{4}\theta}{y^{4}}=\frac{97\sin 2\theta}{x^{3}y+y^{3}x}$

$x^{4}y^{4}(\frac{\cos^{4}\theta}{x^{8}y^{4}}+\frac{\sin^{4}\theta}{y^{8}x^{4}})=\frac{97(2)(\sin \theta)(\cos\theta)}{xy(x^{2}+y^{2})}$

$x^{4}y^{4}(\frac{\sin^{4}\theta}{x^{8}x^{4}}+\frac{\sin^{4}\theta}{y^{8}x^{4}})=\frac{97(2)(\sin \theta)}{x(x^{2}+y^{2})}.\frac{\cos\theta}{y}$

$\sin^{4}\theta(\frac{y^{4}}{x^{8}}+\frac{1}{y^{4}})=\frac{97(2)\sin^{2} \theta}{x^{2}(x^{2}+y^{2})}$

$\sin^{2}\theta(\frac{y^{8}+x^{8}}{x^{8}y^{4}})=\frac{97(2)}{x^{2}(x^{2}+y^{2})}$

$\sin^{2}\theta=\frac{97(2)x^{6}y^{4}}{(x^{2}+y^{2})(x^{8}+y^{8})}$

Similarly,

$\cos^{2}\theta=\frac{97(2)x^{4}y^{6}}{(x^{2}+y^{2})(x^{8}+y^{8})}$

We know,

$\sin^{2}\theta+\cos^{2}\theta=1$

Placing values,

$\frac{97(2)x^{6}y^{4}}{(x^{2}+y^{2})(x^{8}+y^{8})}+\frac{97(2)x^{4}y^{6}}{(x^{2}+y^{2})(x^{8}+y^{8})}=1$

$\frac{97(2)x^{4}y^{4}(x^{2}+y^{2})}{(x^{2}+y^{2})(x^{8}+y^{8})}=1$

$97(2)x^{4}y^{4}=x^{8}+y^{8}$

$\frac{x^{4}}{y^{4}}+\frac{y^{4}}{x^{4}}=194$

$\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}=14$

$\frac{x}{y}+\frac{y}{x}=4$

Truly, the problem is beautiful and would be a delight for algebra lovers.

$\large \frac{\sin\theta}{x} =\frac{\cos\theta}{y} = k$

Then, $\large \sin \theta = xk , \cos \theta = yk$ .

Note that $\large \sin^2 \theta \ + \cos^2 \theta$ = 1. This implies, $\large (xk)^2 + (yk)^2 = 1$ We then obtain, $\large x^2 + y^2 = \frac{1}{k^2}$ .

We now substitute the values of $\large \sin \theta , \cos \theta$ in the given equation.

On simplifying, we obtain $\large k^2( \frac{x^4}{y^4} + \frac{y^4}{x^4})$ = $\frac{194}{x^2+y^2}$ .

Putting the value of $\large k^2$ in the above equation, we obtain $\large ( \frac{x^4}{y^4} + \frac{y^4}{x^4})$ = $\large 194$ .

Put $\large \frac{x^4}{y^4} = \tan^4 \theta = a , \frac{y^4}{x^4}= \cot^4 \theta = b$ . Note that ab =1.

At this stage, we have $\large a^4 + b^4=194$ . On observation, we notice that we have to calculate a + b.

Writing $\large (a^4+b^4) = {((a+b)^2 -2ab})^2 - 2(a^2)(b^2) =194$ and putting the value of ab=1 , we obtain $\large ((a+b)^2 -2)^2 = 196$

Put $\large (a+b)^2$ for T, expand the equation and solve the quadratic so obtained to get the required positive value of T.