To answer try answer by your own are expected to know the following:

Chain Rule:

if #f(x)=g(h(x))# then #f'(x) =g'(h(x)) h'(x) #

Product Rule:

if #[f(x)g(x)]'=f'(x) g(x)+f(x)g'(x) #

The fact that:

#[d/dx]e^x=e^x#

And #[d/dx]sec(x)=sec(x)tan(x)#

Solving:

#d/dx[-1]=d/dx[xy^2+x^2y-e^y sec(xy)]#

It helps to split:

(1)

#d/dx[xy^2]=(x)(d/dx[y^2])+(1)(y^2)=2xyy'+y^2#

(2)

#d/dx[x^2 y]=(x^2)(d/dx[y])+(2x)(y) =x^2(y')(1)+2xy=x^2y'+2xy#

(3a) Before proceeding, it is better to solve #d/dx[sec(xy)]#

Since you need to product rule the 3rd term, then chain rule #sec(xy)# then product rule again #xy# which can get messy.

#d/dx[sec(xy)]=(sec(xy)tan(xy))(d/dx[xy])=sec(xy)tan(xy)(xy'+y)#

(3b)

#d/dx[e^y sec(xy)]=e^y(d/dx[sec(xy)])+(d/dx[e^y])(sec(xy))=e^y(sec(xy)tan(xy)(xy'+y))+(e^yy')(sec(xy))=e^yxy'sec(xy)tan(xy)+
e^yysec(xy)tan(xy)+(e^yy')sec(xy)#

Now putting it all together

#d/dx[-1]=d/dx[xy^2+x^2y-e^y sec(xy)]#

#0=[2xyy'+y^2]+[x^2y'+2xy]-[e^yxy'sec(xy)tan(xy)+
e^yysec(xy)tan(xy)+(e^yy')sec(xy)]#

Put all terms with #y'# on one side and those without in the other.

#0=[2xyy'+y^2]+[x^2y'+2xy]-[e^yxy'sec(xy)tan(xy)+
e^yysec(xy)tan(xy)+e^yy'sec(xy)]#

#-2xyy'-x^2y'+e^yxy'sec(xy)tan(xy)+e^yy'sec(xy)=y^2+2xy-e^yysec(xy)tan(xy)#

#y'(-2xy-x^2+e^yxsec(xy)tan(xy)+e^ysec(xy))=y^2+2xy-e^yysec(xy)tan(xy)#

#y'=(y^2+2xy-e^yysec(xy)tan(xy))/(e^yxsec(xy)tan(xy)+e^ysec(xy)-2xy-x^2)#

Tips: It helps to break down terms and solve each. Be mindful of parantheses And watch out for rules to apply especially if there functions within a function.