sqrt(x)/y^2+4+15/8=2x⇒8sqrt(x)+47y^2=16xy^2⇒y^2(16x-47)=8sqrt(x), y^2=8sqrt(x)/(16x-47).
2ydy/dx=(4(16x-47)/sqrt(x)-128sqrt(x))/(16x-47)^2 is the tangent equation.
At (1,2) this becomes: 4dy/dx=(4(-31)-128)/961, dy/dx=-252/3844.
The equation of the tangent is y=mx+b where m=-63/961; y=-63x/961+b. Since this passes through (1,2), 2=-63/961+b and b=1985/961. So the equation of the tangent at (1,2) is y=(1985-63x)/961.