I think you added an extra #y# and #e^x# here by mistake, because the equation you provided does not contain the point #(1,1)#. However, #xy^2+e^y=2x^2-y+e^x# does contain #(1,1)#, so I'm assuming you meant that.
We start by taking the derivative with respect to #x# on both sides - this is the first step in finding the slope of the tangent line:
#d/dx(xy^2+e^y)=d/dx(2x^2-y+e^x#)
Left Side
We're trying to find #d/dx(xy^2+e^y)#. The sum rule says we can break this up into: #d/dx(xy^2)+d/dx(e^y)#. Starting with #d/dx(xy^2)#:
#color(white)(XX)d/dx(xy^2)=(x)'(y^2)+(x)(y^2)'->#using product rule
#color(white)(XX)d/dx(xy^2)=y^2+2xydy/dx#
Onto #d/dx(e^y)#:
#color(white)(XX)d/dx(e^y)=e^ydy/dx#
Therefore #d/dx(xy^2+e^y)=y^2+2xydy/dx+e^ydy/dx#. That makes our problem come down to:
#y^2+2xydy/dx+e^ydy/dx=d/dx(2x^2-y+e^x)#
Right Side
We're trying to find #d/dx(2x^2-y+e^x)#. Again, we can use the sum rule to break it down into: #d/dx(2x^2)-d/dx(y)+d/dx(e^x)#. These derivatives aren't too hard to manage:
#color(white)(XX)d/dx(2x^2)=4x#
#color(white)(XX)d/dx(y)=dy/dx#
#color(white)(XX)d/dx(e^x)=e^x#
Therefore #d/dx(2x^2-y+e^x)=4x-dy/dx+e^x#. That means our problem is:
#y^2+2xydy/dx+e^ydy/dx=4x-dy/dx+e^x#
Rearranging to solve for #dy/dx#, we see:
#y^2+2xydy/dx+e^ydy/dx=4x-dy/dx+e^x#
#2xydy/dx+e^ydy/dx+dy/dx=4x+e^x-y^2#
#dy/dx(2xy+e^y+1)=4x+e^x-y^2#
#dy/dx=(4x+e^x-y^2)/(2xy+e^y+1)#
Therefore the slope of the tangent line is given by the expression: #(4x+e^x-y^2)/(2xy+e^y+1)#. We're trying to find the equation, which means finding the slope; and to do that, we simply find #dy/dx# at #(1,1)#:
#dy/dx=(4x+e^x-y^2)/(2xy+e^y+1)#
#dy/dx=(4(1)+e^((1))-(1)^2)/(2(1)(1)+e^((1))+1)#
#dy/dx=(3+e^(1))/(3+e^(1))=1#
Thus the slope of the tangent line is #1#. The equation will be of the form: #y=mx+b#, where #x# and #y# are points on the line, #m# is the slope, and #b# is the #y#-intercept. We know everything except #b#, but we have everything except #b#; that means we can use #x#, #y#, and #m# to find #b#:
#y=mx+b#
#1=(1)(1)+b#
#0=b#
That means the equation of the tangent line is #y=x#.
