For the curve $$C: (x^2 + y^2 - 3) + (x^2 - y^2 - 1)^{5} = 0$$, the value of $$3y' - y^3 y''$$, at the point $$(\alpha, \alpha)$$, $$\alpha > 0$$ on C, is equal to ________.

Given curve,

$$x^2+y^2-3+(x^2-y^2-1)^5=0$$

First find the point $$(\alpha,\alpha).$$

Substituting

$$x=\alpha,\qquad y=\alpha,$$

we get

$$\alpha^2+\alpha^2-3+(\alpha^2-\alpha^2-1)^5=0$$

$$2\alpha^2-3+(-1)^5=0$$

$$2\alpha^2-4=0$$

$$\alpha^2=2$$

Since $$\alpha>0,$$

$$\alpha=\sqrt2$$

Hence, the point is

$$(\sqrt2,\sqrt2)$$

Now differentiate implicitly:

$$2x+2yy'+5(x^2-y^2-1)^4(2x-2yy')=0$$

Dividing by $$2,$$

$$x+yy'+5(x^2-y^2-1)^4(x-yy')=0$$

Substituting

$$x=y=\sqrt2,$$

and using

$$x^2-y^2-1=-1,$$

we get

$$\sqrt2+\sqrt2\,y'+5(\sqrt2-\sqrt2\,y')=0$$

$$1+y'+5-5y'=0$$

$$4y'=6$$

$$y'=\frac32$$

Now differentiate

$$x+yy'+5(x^2-y^2-1)^4(x-yy')=0$$

with respect to $$x$$:

$$1+(y')^2+yy''+5\left[4(x^2-y^2-1)^3(2x-2yy')(x-yy')+(x^2-y^2-1)^4(1-(y')^2-yy'')\right]=0$$

Substituting

$$x=y=\sqrt2,\qquad y'=\frac32,\qquad x^2-y^2-1=-1,$$

we get

$$1+\frac94+\sqrt2\,y''+5\left[4(-1)^3(2\sqrt2-3\sqrt2)\left(\sqrt2-\frac{3\sqrt2}{2}\right)+(-1)^4\left(1-\frac94-\sqrt2\,y''\right)\right]=0$$

$$\frac{13}{4}+\sqrt2\,y''+5\left[4(-1)(-\sqrt2)\left(-\frac{\sqrt2}{2}\right)-\frac54-\sqrt2\,y''\right]=0$$

$$\frac{13}{4}+\sqrt2\,y''+5\left[-4-\frac54-\sqrt2\,y''\right]=0$$

$$\frac{13}{4}+\sqrt2\,y''-\frac{105}{4}-5\sqrt2\,y''=0$$

$$-23-4\sqrt2\,y''=0$$

$$y''=-\frac{23}{4\sqrt2}$$

Now compute

$$3y'-y^3y''$$

$$=3\left(\frac32\right)-(\sqrt2)^3\left(-\frac{23}{4\sqrt2}\right)$$

$$=\frac92+\frac{23}{2}$$

$$=16$$

Hence, the required value is

$$\boxed{16}$$.