Find y' ?? please help

#sqrt[4x^2+y]#.... =#4x^6e^[2y]# squaring both sides will give
#4x^2+y#=# 4x^6e^[2y]#........Differentiating both sides
implicitly with respect #x# and using the product rule on the right hand side, i.e, #d[uv]=vdu+udv#

Let #u =4x^6 and# #v=e^[2y]#

#8x+dy/dx#=#e^[2y]d/dx4x^6#+#4x^6d/dxe^[2y]#

So , #8x+dy/dx#=#e^[2y]##24x^5#+#4x^6##e^[2y]##.2dy/dx# ,rearranging and factoring........

#dy/dx-8x^6e^[2y]dy/dx#=#24x^5e^[2y]-8x#

#dy/dx[1-8x^6e^[2y]]#=#24x^5e^[2y]-8x# and so,

#y'#=#[24x^5e^[2y]-8x]/[1-8x^6e^[2y]#

We have:

# sqrt(4x^2+y) = 2x^3e^y #

If we have a function #F(x, y) = 0# which defines #y# as a function of #x, y = y(x)#, then:

# dy/dx = − F_x / F_y #

So Let # F(x,y) = sqrt(4x^2+y)-2x^3e^y #; Then;

Then:

# F_x = (partial F)/(partial x) #

# \ \ \ \ = 1/2(4x^2+y)^(-1/2)(8x)-6x^2e^y #

# \ \ \ \ = (4x)/sqrt(4x^2+y)-6x^2e^y #

And:

# F_y = (partial F)/(partial y) #

# \ \ \ \ = 1/2(4x^2+y)^(-1/2)(1)-2x^3e^y #

# \ \ \ \ = 1/(2sqrt(4x^2+y))-2x^3e^y #

And so we have:

# dy/dx = − F_x / F_y #

# \ \ \ \ \ = − ((4x)/sqrt(4x^2+y)-6x^2e^y) / (1/(2sqrt(4x^2+y))-2x^3e^y) #

Using the original definition:

# sqrt(4x^2+y) = 2x^3e^y #

We can simplify our expression:

# dy/dx = − ((4x)/(2x^3e^y)-6x^2e^y) / (1/(2(2x^3e^y))-2x^3e^y) #

# \ \ \ \ \ = − ( ( 4x -6x^2e^y(2x^3e^y) ) / (2x^3e^y) ) / ( ( 1-2x^3e^y(2)(2x^3e^y) ) / ((2)(2x^3e^y) ) #

# \ \ \ \ \ = −2 ( 4x -6x^2e^y(2x^3e^y) ) / ( 1-2x^3e^y(4x^3e^y) ) #

# \ \ \ \ \ = ( 24x^5e^2y -8x) / ( 1- 8x^6e^(2y) ) #